All editorial codes contain solutions wrapped in classes. I'll only paste the important part (namespace solution in C++, class Solution in Python 3) as plaintext in the solution code section (along with the corresponding AC submission link). You might want to see the code template here.
#include <bits/stdc++.h>
using namespace std;
namespace solution {
bool hasMultipleTests = true;
void preprocess() {
// something
}
void input(int testcase) {
// something
}
void solve(int testcase) {
// something
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
solution::preprocess();
int t = 1;
if (solution::hasMultipleTests) cin >> t;
for (int testcase=1; testcase<=t; testcase++) {
solution::input(testcase);
solution::solve(testcase);
}
return 0;
}
import sys
input = sys.stdin.readline
class Solution:
hasMultipleTests = True
@classmethod
def preprocess(cls):
pass
@classmethod
def input(cls, testcase):
pass
@classmethod
def solve(cls, testcase):
pass
# end Solution
if __name__ == '__main__':
Solution.preprocess()
t = int(input()) if Solution.hasMultipleTests else 1
for testcase in range(1, t+1):
Solution.input(testcase)
Solution.solve(testcase)
Also, per my own custom ever since 2019, solution codes will only be published after system testing, to ensure all codes used here guarantee AC. Thank you for your patience.
Author: AkiLotus
Development: AkiLotus
Hinter: AkiLotus
Editorialist: AkiLotus, Kuroni
In which cases would a light be on?
Imagine that you have $$$x$$$ white socks and $$$y$$$ black socks. What would you do to make the maximum pairs of matched socks?
Imagine that you have $$$x$$$ white socks and $$$y$$$ black socks. What would you do to make the maximum pairs of unmatched socks?
Submission link: 289291089
namespace solution {
bool hasMultipleTests = true;
int n;
vector<int> a;
void preprocess() {
}
void input(int testcase) {
cin >> n;
a.clear();
a.resize(n * 2);
for (auto &z: a) cin >> z;
}
void solve(int testcase) {
int cnt0 = 0;
for (auto &z: a) {
cnt0 += z;
}
cout << (cnt0 & 1) << " " << min(cnt0, n * 2 - cnt0) << endl;
}
}
Submission link: 289291082
class Solution:
hasMultipleTests = True
n: int = None
a: list = None
@classmethod
def preprocess(cls):
pass
@classmethod
def input(cls, testcase):
cls.n = int(input())
cls.a = list(map(int, input().split()))
@classmethod
def solve(cls, testcase):
cnt0 = sum(cls.a)
print(cnt0 & 1, min(cnt0, cls.n*2 - cnt0))
# end Solution
Author: Kuroni
Development: AkiLotus
Hinter: Kuroni
Editorialist: AkiLotus, Kuroni
There is always exactly one solution for $$$n = 1$$$. For $$$n \ge 3$$$, for which $$$k$$$ should there be no solution at all?
Try to find a solution with $$$k = 2$$$ and $$$k = 4$$$, assuming $$$n$$$ allows at least one solution to exist. Can you find a pattern and generalize it for all even $$$k$$$?
Try to find a solution with $$$k = 3$$$ and $$$k = 5$$$, assuming $$$n$$$ allows at least one solution to exist. Can you find a pattern and generalize it for all odd $$$k$$$?
Submission link: 289291698
namespace solution {
bool hasMultipleTests = true;
int n, k;
void preprocess() {
}
void input(int testcase) {
cin >> n >> k;
}
void solve(int testcase) {
if (n == 1) {cout << "1\n1\n"; return;}
if (k == 1 || k == n) {cout << "-1\n"; return;}
int p2 = k - k % 2;
int p3 = k + 1 + k % 2;
cout << "3\n1 " << p2 << " " << p3 << endl;
}
}
Submission link: 289291693
class Solution:
hasMultipleTests = True
n: int = None
k: int = None
@classmethod
def preprocess(cls):
pass
@classmethod
def input(cls, testcase):
cls.n, cls.k = map(int, input().split())
@classmethod
def solve(cls, testcase):
if cls.n == 1: return(print('1\n1'))
if cls.k in {1, cls.n}: return(print(-1))
p2, p3 = cls.k - cls.k % 2, cls.k + 1 + cls.k % 2
print(f'3\n1 {p2} {p3}')
# end Solution
Author: AkiLotus
Development: AkiLotus
Hinter: AkiLotus
Editorialist: AkiLotus, Kuroni
Without loss of generality, you can rewrite the triangle inequality into a much more compacted form if the side lengths are properly ordered.
Given four integers $$$a \le b \le c \le d$$$, if $$$(a, b, d)$$$ are side lengths of a non-degenerate triangle, so does $$$(a, b, c)$$$.
Given four integers $$$a \le b \le c \le d$$$, if $$$(a, b, d)$$$ are side lengths of a non-degenerate triangle, so does $$$(a, c, d)$$$.
Given four integers $$$a \le b \le c \le d$$$, if $$$(a, b, d)$$$ are side lengths of a non-degenerate triangle, so does $$$(b, c, d)$$$.
To make a sorted array $$$[b_1, b_2, \ldots, b_m] \space (m \ge 3)$$$ satisfy the problem, we must ensure that $$$b_1 + b_2 > b_m$$$.
Fix the result array to have two minimas and one maxima at respectively $$$a_x$$$, $$$a_y$$$ and $$$a_z$$$. How many operations are required to do so?
Is there a fast and optimal way to try all possible $$$(a_x, a_y, a_z)$$$ triplets?
Submission link: 289292343
namespace solution {
bool hasMultipleTests = true;
int n;
vector<int> a;
void preprocess() {
}
void input(int testcase) {
cin >> n;
a.clear();
a.resize(n);
for (auto &z: a) cin >> z;
}
void solve(int testcase) {
sort(a.begin(), a.end());
int l = 0, ans = n - 2;
for (int r = 2; r < n; r++) {
while (r - l >= 2 && a[l] + a[l+1] <= a[r]) l++;
ans = min(ans, n - (r - l + 1));
}
cout << ans << endl;
}
}
Submission link: 289292338
class Solution:
hasMultipleTests = True
n: int = None
a: list = None
@classmethod
def preprocess(cls):
pass
@classmethod
def input(cls, testcase):
cls.n = int(input())
cls.a = list(map(int, input().split()))
@classmethod
def solve(cls, testcase):
cls.a.sort()
l, r, ans = 0, 2, cls.n - 2
while r < cls.n:
while r - l >= 2 and cls.a[l] + cls.a[l+1] <= cls.a[r]: l += 1
ans = min(ans, cls.n - (r - l + 1))
r += 1
print(ans)
# end Solution
Author: AkiLotus
Development: AkiLotus
Hinter: AkiLotus
Editorialist: AkiLotus, Kuroni
There are many specific constraints on the system tree.
First, notice that $$$1 \le x \le y \le n-1$$$ if and only if $$$p_x \le p_y$$$. Draw a few trees with this constraint and start pointing your finger on the nodes from $$$0$$$ to $$$n-1$$$. Notice the pattern your finger is making.
Incorporate hint #1 with the fact that the "tentacles" of the Genokraken are just paths. Try to draw more detailed conclusion about that pattern.
Is there a way to know how many tentacles there are?
What use could we make of the fact that node $$$1$$$ has exactly two adjacent nodes?
The query limit is very small. Is there a way to query that the information obtained in each query should (at least partially) determine a node?
For each "tentacle", can we know when its path has ended? How many queries should it take?
Submission link: 289292866
namespace datastruct {
// A doubly-linked list that only supports modify and (pseudo) deletion at pointer
class DeleteOnly_DLL {
public:
vector<int> values;
vector<int> prev;
vector<int> next;
int pointer;
DeleteOnly_DLL(int size) {
values.resize(size);
prev.resize(size);
next.resize(size);
for (int i = 0; i < size; i++) {
prev[i] = (i + size - 1) % size;
next[i] = (i + 1) % size;
}
pointer = 0;
}
int current() {
return values[pointer];
}
// Set value at pointer and move pointer to next
void set_and_move(int val) {
values[pointer] = val;
pointer = next[pointer];
}
// "Delete" node and move pointer to next
void erase() {
if (prev[pointer] != -1) {
next[prev[pointer]] = next[pointer];
}
if (next[pointer] != -1) {
prev[next[pointer]] = prev[pointer];
}
int next_id = next[pointer];
prev[pointer] = next[pointer] = -1;
pointer = next_id;
}
};
}
namespace solution {
bool hasMultipleTests = true;
int n;
int ask(int a, int b) {
cout << "? " << a << " " << b << endl;
cout.flush();
int res; cin >> res;
return res;
}
void answer(vector<int> &p) {
cout << "!";
for (int i=1; i<n; i++) {
cout << " " << p[i];
}
cout << endl;
cout.flush();
}
void preprocess() {
}
void input(int testcase) {
cin >> n;
}
void solve(int testcase) {
vector<int> p(n, -1);
p[1] = 0;
int r = 2;
while (true) {
int response = ask(1, r);
if (response == -1) exit(2226);
if (response == 1) {
p[r] = 0;
r++;
}
else break;
}
int tentacle_count = r - 1;
datastruct::DeleteOnly_DLL tentacles = datastruct::DeleteOnly_DLL(tentacle_count);
for (int i = 0; i < tentacle_count; i++) {
tentacles.set_and_move(i + 1);
}
p[r] = tentacles.current();
tentacles.set_and_move(r);
r++;
while (r < n) {
int response = ask(tentacles.current(), r);
if (response == -1) exit(2226);
if (response == 1) {
tentacles.erase();
}
else {
p[r] = tentacles.current();
tentacles.set_and_move(r);
r++;
}
}
answer(p);
}
}
Submission link: 289292865
class DeleteOnly_DLL:
def __init__(self, size: int):
self.values = [None for _ in range(size)]
self.prev = [(i + size - 1) % size for i in range(size)]
self.next = [(i + 1) % size for i in range(size)]
self.pointer = 0
def current(self):
return self.values[self.pointer]
# Set value at pointer and move pointer to next
def set_and_move(self, val):
self.values[self.pointer] = val
self.pointer = self.next[self.pointer]
# "Delete" node and move pointer to next
def erase(self):
if self.prev[self.pointer] != -1:
self.next[self.prev[self.pointer]] = self.next[self.pointer]
if self.next[self.pointer] != -1:
self.prev[self.next[self.pointer]] = self.prev[self.pointer]
next_id = self.next[self.pointer]
self.prev[self.pointer] = self.next[self.pointer] = -1
self.pointer = next_id
# end DeleteOnly_DLL
class Solution:
hasMultipleTests = True
n: int = None
@classmethod
def ask(cls, a: int, b: int):
print(f'? {a} {b}', flush=True)
return int(input())
@classmethod
def answer(cls, p: list):
print(f'! {" ".join(map(str, p[1:]))}', flush=True)
@classmethod
def preprocess(cls):
pass
@classmethod
def input(cls, testcase):
cls.n = int(input())
@classmethod
def solve(cls, testcase):
p = [-1 for _ in range(cls.n)]
p[1] = 0
r = 2
while True:
response = cls.ask(1, r)
if response == -1: exit(2226)
if response == 1:
p[r] = 0
r += 1
else: break
tentacle_count = r - 1
tentacles = DeleteOnly_DLL(size = tentacle_count)
for i in range(tentacle_count):
tentacles.set_and_move(i + 1)
p[r] = tentacles.current()
tentacles.set_and_move(r)
r += 1
while r < cls.n:
response = cls.ask(tentacles.current(), r)
if response == -1: exit(2226)
if response == 1:
tentacles.erase()
else:
p[r] = tentacles.current()
tentacles.set_and_move(r)
r += 1
cls.answer(p)
# end Solution
Author: thanhchauns2
Development: thanhchauns2, Sagita_Phoenix
Hinter: AkiLotus
Editorialist: Kuroni, AkiLotus
Why is $$$n$$$ odd? Does it imply anything of the availability of a solution?
$$$n = 1$$$ is trivial. Try to solve this problem if $$$n = 3$$$ and $$$a_1 = a_3$$$.
Try to solve this problem for any array with $$$n = 3$$$. Is there any way to solve this without going pass a state of $$$a_1 = a_3$$$ (or $$$a_1 = a_2$$$ or $$$a_2 = a_3$$$)?
Assume that $$$n = 5$$$, $$$a_1 = a_5$$$ and $$$a_2 = a_3 = a_4$$$. Now refer to hint #2, can you see any similarity in the solution?
Refer to hint #4. What would happen if we apply an operation to index $$$2$$$ and another to index $$$4$$$?
For any array with $$$n = 5$$$, how would you convert it into the form stated in hint #4?
Refer to hint #6. How would you make $$$a_1 = a_5$$$ and $$$a_2 = a_4$$$? Does the fact that $$$n$$$ is odd help in any way?
Can you draw a generalized conclusion?
Submission link: 289293300
namespace solution {
bool hasMultipleTests = true;
int n;
vector<int> a;
void apply_prefixes(vector<long long> &prefixes, vector<long long> &v) {
for (int i = 2; i < n * 2; i++) {
prefixes[i] += prefixes[i - 2];
}
for (int i = 0; i < n * 2; i++) {
v[i % n] += prefixes[i];
}
}
void construct_trench(vector<long long> &arr, vector<long long> &v) {
vector<long long> prefixes(n * 2, 0LL);
vector<long long> delta(n, 0LL);
for (int i = 0; i < n / 2; i++) {
long long diff = arr[n - 1 - i] - (arr[i] + delta[i]);
delta[i] += 2 * diff;
delta[i + 1] += diff;
prefixes[n - i] += diff;
prefixes[n + i + 2] -= diff;
}
apply_prefixes(prefixes, v);
for (int i = 0; i < n; i++) {
arr[i] += v[i] * 2;
arr[(i + 1) % n] += v[i];
arr[(i + n - 1) % n] += v[i];
}
}
void construct_plateau(vector<long long> &arr, vector<long long> &v) {
vector<long long> prefixes(n * 2, 0LL);
vector<long long> delta(n, 0LL);
for (int i = n / 2 - 1; i >= 0; i--) {
long long diff = arr[i] - (arr[i + 1] + delta[i + 1]);
delta[i] += diff;
prefixes[i + 1] += diff;
prefixes[n - i] -= diff;
}
apply_prefixes(prefixes, v);
}
void preprocess() {
}
void input(int testcase) {
cin >> n;
a.clear();
a.resize(n);
for (auto &z: a) cin >> z;
}
void solve(int testcase) {
if (n == 1) {
cout << "0\n";
return;
}
vector<long long> v(n, 0LL);
vector<long long> arr(n);
for (int i = 0; i < n; i++) {
arr[i] = 1LL * a[i];
}
construct_trench(arr, v);
construct_plateau(arr, v);
long long offset = *ranges::min_element(v);
for (auto &z: v) cout << z - offset << " ";
cout << endl;
}
}
Submission link: 289293299
class Solution:
hasMultipleTests = True
n: int = None
a: list = None
@classmethod
def apply_prefixes(cls, prefixes, v):
for i in range(2, cls.n*2):
prefixes[i] += prefixes[i - 2]
for i in range(cls.n*2):
v[i % cls.n] += prefixes[i]
@classmethod
def construct_trench(cls, arr, v):
prefixes = [0 for _ in range(cls.n * 2)]
delta = [0 for _ in range(cls.n)]
for i in range(cls.n // 2):
diff = arr[cls.n - 1 - i] - (arr[i] + delta[i])
delta[i] += 2 * diff
delta[i + 1] += diff
prefixes[cls.n - i] += diff
prefixes[cls.n + i + 2] -= diff
cls.apply_prefixes(prefixes, v)
for i in range(cls.n):
arr[i] += v[i] * 2
arr[(i + 1) % cls.n] += v[i]
arr[(i + cls.n - 1) % cls.n] += v[i]
@classmethod
def construct_plateau(cls, arr, v):
prefixes = [0 for _ in range(cls.n * 2)]
delta = [0 for _ in range(cls.n)]
for i in range(cls.n // 2 - 1, -1, -1):
diff = arr[i] - (arr[i + 1] + delta[i + 1])
delta[i] += diff
prefixes[i + 1] += diff
prefixes[cls.n - i] -= diff
cls.apply_prefixes(prefixes, v)
@classmethod
def preprocess(cls):
pass
@classmethod
def input(cls, testcase):
cls.n = int(input())
cls.a = list(map(int, input().split()))
@classmethod
def solve(cls, testcase):
if cls.n == 1:
return(print(0))
v = [0 for _ in range(cls.n)]
cls.construct_trench(cls.a, v)
cls.construct_plateau(cls.a, v)
offset = min(v)
v = list(map(lambda x: x - offset, v))
print(*v)
# end Solution
We have some observations here. Our only allowed operation is $$$[1, 2, 1]$$$, but by stacking it in the right way (and normalizing the differences) we could obtain simpler operations. In details:
- We can apply operation $$$[-1, -1]$$$ to index $$$i$$$ (decrease values at index $$$i$$$ and $$$i+1$$$ by $$$1$$$) by applying operation $$$[1, 2, 1]$$$ to index $$$i+2$$$, $$$i+4$$$, and so on until $$$i-1$$$ (please treat these addition/subtraction of indices in a cyclic array context). Doing so will increase every index's value by $$$2$$$, except $$$i$$$ and $$$i+1$$$ only get increase by $$$1$$$, so relatively, indices $$$i$$$ and $$$i+1$$$ got their values lowered by $$$1$$$ compared to the rest.
- We can apply operation $$$[1]$$$ to index $$$i$$$ (increase value at index $$$i$$$ by $$$1$$$) by applying operation $$$[-1, -1]$$$ to index $$$i+1$$$, $$$i+3$$$, and so on until $$$i-1$$$. Proof for this is trivial.
Author: MofK
Development: MofK, AkiLotus
Hinter: AkiLotus
Editorialist: MofK, AkiLotus
Try to solve if all pockets are in the same box. Does the problem seem familiar?
There's one kind of cases for a box that its game has a fixed move sequence, and thus a fixed outcome. What is it?
Let's take the cases found in hint #2 out of the equation from here on. Is there any way for a player to win in two consecutive boxes?
If a player can win a box if both play optimally to win, can they also lose it if both play optimally to lose?
Submission link: 289293663
namespace solution {
bool hasMultipleTests = true;
int n;
vector<int> a;
const int MAXN = 1000000;
const int MOD = 998244353;
int pow2[MAXN];
void preprocess() {
pow2[0] = 1;
for (int i = 1; i < MAXN; i++) {
pow2[i] = (2LL * pow2[i-1]) % MOD;
}
}
void input(int testcase) {
cin >> n;
a.clear();
a.resize(n);
for (auto &z: a) cin >> z;
}
void solve(int testcase) {
if (*ranges::max_element(a) == 1) {
cout << (n & 1 ? pow2[n-1] : 0) << endl;
return;
}
int ans = 0;
// The critical layer (assuming prefix Grundy > 1) can only fall into Alice's control
// if and only if before it is an even amount of pockets
int alice_at_critical = 1;
int prefix_1 = 0;
while (a[prefix_1] == 1) {
prefix_1++;
if (prefix_1 % 2 == 0) {
alice_at_critical = (alice_at_critical + pow2[prefix_1 - 1]) % MOD;
}
}
int grundy = (prefix_1 & 1);
for (int r = prefix_1; r < n; r++) {
grundy ^= a[r];
if (!grundy) continue;
int post_critical = (r < n - 1 ? pow2[n - 2 - r] : 1);
int pre_critical = (grundy == 1 ? pow2[prefix_1] : alice_at_critical);
ans += (1LL * pre_critical * post_critical) % MOD;
ans %= MOD;
}
cout << ans << endl;
}
}
Submission link: 289293585
class Solution:
hasMultipleTests = True
n: int = None
a: list = None
MAXN: int = 1000000
MOD: int = 998244353
pow2: list = None
@classmethod
def preprocess(cls):
cls.pow2 = [None for _ in range(cls.MAXN)]
for i in range(cls.MAXN):
cls.pow2[i] = 1 if i == 0 else (2 * cls.pow2[i-1]) % cls.MOD
@classmethod
def input(cls, testcase):
cls.n = int(input())
cls.a = list(map(int, input().split()))
@classmethod
def solve(cls, testcase):
if max(cls.a) == 1:
return(print(cls.pow2[cls.n-1] if cls.n & 1 else 0))
ans = 0
# The critical layer (assuming prefix Grundy > 1) can only fall into Alice's control
# if and only if before it is an even amount of pockets
alice_at_critical = 1
prefix_1 = 0
while cls.a[prefix_1] == 1:
prefix_1 += 1
if prefix_1 % 2 == 0:
alice_at_critical = (alice_at_critical + cls.pow2[prefix_1 - 1]) % cls.MOD
grundy = prefix_1 & 1
for r in range(prefix_1, cls.n):
grundy ^= cls.a[r]
if grundy == 0: continue
post_critical = cls.pow2[cls.n - 2 - r] if r < cls.n - 1 else 1
pre_critical = cls.pow2[prefix_1] if grundy == 1 else alice_at_critical
ans = (ans + pre_critical * post_critical) % cls.MOD
print(ans)
# end Solution
Problem D can be solved in $$$\mathcal{O}(n)$$$ with a queue quite easily as well, which is implemented directly into the stl so it's easier. The idea stays the same, just replace the double linked list with a queue
I believe there is an even lighter ( implementation wise ) sol to D. I keep track of the next parent to be considered ( stored in variable bp initiated at N-2 ) and the current node ind ( initiated at N-1) who's parent we're looking for, we (decrement ind iff bp is its parent) and decrement bp at each iteration till its 0
That's quite brilliant, though it feels less intuitive to me
Tutorial of D definitely proves that the problem setter group had Asian influence ..
W contest.
For D, there also exists n — 3 query soln instead of 2 * n — 6 query solution.
can someone pls prove the fact that the smallest two elements will always have to be continuous in the original array as well ? I am refering to problem C here
what do you mean by continuous?
another solution to E, although after reading the alternate solution, I think that is better
we can apply operation [-1,-1,1,1] to index i (e.g. a[i]--,a[i+1]--,a[i+2]++,a[i+3]++) by applying operation [-1,-2,-1] to i, and [1,2,1] to i+1
now, we can apply operation [-1,0,1] to index i by applying operation [-1,-1,1,1] to indexes i,i+2,...,until i-1
we can apply operation [-1,1] to index i by applying operation [-1,0,1] to indexes i,i+2,... until i-1
now applying operation [-1,1] keeps the sum constant, so we just need apply arbitrary operations [1,2,1] until (sum of a)%n==0 which is always possible
289298093
Alternate solutions for E
Write the equations A[i+1] + V[i] + 2 * V[i+1] + V[i+2] = T. [eqn1] Taking differences, (A[i+2] — A[i+1]) + (V[i+3] + V[i+2]) — (V[i+1] + V[i]) = 0 [eqn2]
Solution 1. https://codeforces.net/contest/2032/submission/289289897 Write W[i] = V[i] + V[i+1]. eqn2 is W[i+2] = W[i] + A[i+1] — A[i+2], so we can fill all W[i]'s in terms of W[0], and all coefficients of W[0] are 1. So we can sum all rows of eqn1 to find W[0]. Now we want to solve for V in terms of W, which is easy by taking alternating sums of V[i+1] = W[i] — V[i].
Solution 2. https://codeforces.net/contest/2032/submission/289297804 Notice a solution for T=0 implies a solution for T=4k (V[i] += k). So, guarantee a solution for T=0 by first shifting A, then we can assume T=0. Now V[0] and V[1] determine V. By chasing eqn1, we will get 2 equations and 2 unknowns.
wice city
For C, I sorted then binary search on the minimum number of elements to reassign. 289211192
Why was n forced odd in E?
in problem C whats the proof that ax and ay should be consecutive elements of the array in sorted order ? suppose we have x as operations needed using consecutive a[i] and a[i+1] lets say max we can reach is j , now if we let ax to be a[i] and ay to be a[i+2] then j can be much more than the original isnt if we have a[i]+a[i+2]>a[i]+a[i+1]? whats the proof that checking with consecutive elements does the job? AkiLotus
Without that condition, you can't ensure that all triplets are valid.
One counterexample: $$$[3, 4, 5, 7]$$$. If you used $$$a_i$$$ and $$$a_{i+2}$$$ for pivoting, you'd conclude that $$$0$$$ operations are required, which is not true, as $$$(3, 4, 7)$$$ are invalid.
For your example i will consider the array two minimums be 3,5 now j will reach 7 so we have 3,5,7 max so total ops needed is 4-3=1 because i will change the one (i have not taken) like 4 between 3 and 5 to ay that is 5, so that validity is maintained.
I see, you wanna pick $$$a_i$$$ and the range $$$[a_{i+2}, a_j]$$$. Then... why don't you simply pivot $$$a_{i+1}$$$ and $$$a_{i+2}$$$ for a clearly not-lower upper bound $$$a_{j'}$$$?
Finally Pupil