#include <bits/stdc++.h>
using ll = long long;
using dl = double long;
using namespace std;
const ll MOD = 1e9 + 7;

//From index 0 to i, what is the max value we can get with weight = w remaining?
ll f(int i, int w, vector<ll>& wt, vector<ll>& val, vector<vector<ll>>& dp){
    if(i == 0){
        if(wt[0] <= w) return val[0];
        else return 0;
    }
    if(dp[i][w] != -1) return dp[i][w];

    ll notTake = 0 + f(i - 1, w, wt, val, dp);
    ll take = INT_MIN;
    if(wt[i] <= w) take = val[i] + f(i - 1, w - wt[i], wt, val, dp);

    return dp[i][w] = max(take, notTake);
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    
    int n, W;
    cin >> n >> W;
    vector<ll> wt(n);
    vector<ll> val(n);
    for(int i = 0; i < n; i++){
        cin >> wt[i] >> val[i];
    }

    vector<vector<ll>> dp(n + 1, vector<ll> (W + 1, -1));
    ll ans = f(n - 1, W, wt, val, dp);

    cout << ans << endl;

    return 0;
}

#include <bits/stdc++.h>
using ll = long long;
using dl = double long;
using namespace std;
const ll MOD = 1e9 + 7;

//From index 0 to i, to achieve value = v what is the min weight required?
ll f(ll i, ll v, vector<ll>& wt, vector<ll>& val, vector<vector<ll>>& dp){
    if(i == 0){
        if(v == 0) return 0;
        if(val[0] == v) return wt[0];
        else return 1e15;
    }
    if(dp[i][v] != -1) return dp[i][v];

    ll notTake = 0 + f(i - 1, v, wt, val, dp);
    ll take = INT_MAX;
    if(val[i] <= v) take = wt[i] + f(i - 1, v - val[i], wt, val, dp);

    return dp[i][v] = min(take, notTake);
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    
    ll n, W;
    cin >> n >> W;
    vector<ll> wt(n);
    vector<ll> val(n);
    int V = 0;
    for(int i = 0; i < n; i++){
        cin >> wt[i] >> val[i];

        V += val[i];
    }

    vector<vector<ll>> dp(n + 1, vector<ll> (V + 1, -1));

    for(ll v = V; v >= 0; v--){
        if(f(n - 1, v, wt, val, dp) <= W){
            cout << v << endl; //First value from maximum which can be possible within given max. wt(W) of knapsack
            break;
        }
    }

    return 0;
}

#include <bits/stdc++.h>
using ll = long long;
using dl = double long;
using namespace std;
const ll MOD = 1e9 + 7;


int main() {
  int t;
  cin >> t;

  while(t--){
    int n, x, y;
    cin >> n >> x >> y;

    if(x == y){
      cout << max((y - 1), (n - x)) << endl;
      continue; 
    }
    if(x < y) swap(x, y);
    // cout << x << "_" << y << endl;

    int rem = n - 2;
    if(y == 1 && x == n){
      cout << rem / 2 + (rem % 2 != 0) << endl;
      continue;
    }
    if((x == n && y > 1) || (y == 1 && x < n)){
      cout << x - y - 2 << endl;
      continue;
    }

    int temp = x - y - 2;
    cout << max((y - 1) + temp, (n - x) + temp) << endl;
  }
  
  return 0;
}

#include <iostream>
#define ll long long
using namespace std;

int main(){
  int t; cin>>t;
  while(t-->0){
    ll n, x, y; cin>>n>>x>>y;
    x--; y--;
    if(x>y){
      swap(x, y);
    }
    if(x==y){
      cout<<max(x, n-1-y)<<endl;
      continue;
    }
    ll ans = 1e9;
    for(int i=x; i<=y; i++){
      ll minix = min((i-x),x);
      ll maxix = max((i-x),x);
      ll miniy = min((y-i-1),(n-1-y));
      ll maxiy = max((y-i-1),(n-1-y));
      ans = min(ans, max(2*minix+maxix, 2*miniy + maxiy));
    }
    cout<<ans<<endl;
  }
  return 0;
}