Simplifying the XOR Update:

Consider a simpler problem. Let's say that all values in arr can be either 0 or 1, and even $$$x$$$ can be only 0 or 1.

Now the sum of range $$$[l, r]$$$ is $$$S$$$, which indicates that there are $$$S$$$ 1s and $$$r — l + 1 — S$$$ 0s in $$$[l, r]$$$.

1. An update with $$$x = 0$$$ won’t change anything.
2. An update with $$$x = 1$$$ would flip 0s and 1s, so there will be $$$r - l + 1 - S $$$ 1s and $$$S$$$ 0s. The new sum is:

$$$r - l + 1 - S $$$

Thus, we can build a segment tree with lazy propagation for this array.

Generalizing for the Full Problem:

The key idea is to notice that each number can be at most 30 bits, and we can just maintain such arrays for each of the 30 bits.

So, we build $$$LOGMAX (=30)$$$ such arrays, and for each query:

1. For sum $$$[l, r]$$$, calculate $$${sum}_0, \text{sum}_1, \ldots, \text{sum}_{29}$$$ for each of the trees. The final sum is:

$$$\text{sum}_0 \times 2^0 + \text{sum}_1 \times 2^1 + \ldots + \text{sum}_{29} \times 2^{29}$$$

1. For the update $$$[l, r]$$$ with XOR $$$x$$$, flip $$$[l, r]$$$ for the $$$k^{th}$$$ tree if the $$$k^{th}$$$ bit of $$$x$$$ is set.

#include <bits/stdc++.h>
 
using namespace std;
 
//#pragma GCC optimize("Ofast,unroll-loops") 
//#pragma GCC target("avx,avx2,avx512,fma") 
 
template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { os << '{'; string sep; for (const T &x : v) os << sep << x, sep = ", "; return os << '}'; }
void dbg_out() { cerr << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { cerr << ' ' << H; dbg_out(T...); }
#ifdef LOCAL
#define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__)
#else
#define dbg(...)
#endif
 
#define ll long long
#define ld long double
 
#define PI 3.1415926535897932384626433832795l 

// -------------------------<rng>------------------------- 
// RANDOM NUMBER GENERATOR
mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count());  
#define SHUF(v) shuffle(all(v), RNG); 
// Use mt19937_64 for 64 bit random numbers.
 
struct LazySegmentTree {
  int n;
  vector<ll> L, R;
  vector<int> val;
  vector<bool> lazy;

  LazySegmentTree(int sz) {
    n = 4 * sz;
    L.resize(n);
    R.resize(n);
    lazy.resize(n);
    val.resize(n);
    n >>= 1;
    init(1, 0, sz - 1);
  }

  void init(int node, int l, int r) {
    if (l > r) {
      return;
    }
    L[node] = l;
    R[node] = r;
    if (l < r) {
      int mid = (l + r) / 2;
      init(2 * node, l, mid);
      init(2 * node + 1, mid + 1, r);
    }
  }

  void flip(int node){
    lazy[node] = lazy[node] ^ 1;
    val[node] = (R[node] - L[node] + 1) - val[node]; // Updating sum for range
  }

  void propagate(int node) {
    if(L[node] < R[node]){
      if(lazy[node]){
        flip(2 * node);
        flip(2 * node + 1);
        lazy[node] = 0;
      }
    }
  }

  void set(int node, int l, int r) {
    if (r < L[node] || R[node] < l) {
      return;
    }
    if (l <= L[node] && R[node] <= r) {
      flip(node);
    } else {
      propagate(node);
      set(2 * node, l, r);
      set(2 * node + 1, l, r);
      val[node] = val[2 * node] + val[2 * node + 1];
    }
  }

  ll get(int node, int l, int r) {
    if (r < L[node] || R[node] < l) {
      return 0; // Out of range
    }
    if (l <= L[node] && R[node] <= r) {
      return val[node];
    }
    propagate(node);
    ll left_sum = get(2 * node, l, r);
    ll right_sum = get(2 * node + 1, l, r);
    return left_sum + right_sum;
  }
};

const int LOGMAX = 30;
 
vector<ll> solve(int n, vector<vector<int>> &queries) {
  vector<LazySegmentTree> tree;
  for(int k = 0; k < LOGMAX; ++k){
    tree.push_back(LazySegmentTree(n + 1));
  }
  vector<ll> res;
  for(vector<int> &query: queries){
    int type = query[0];
    if(type == 1){
      int l = query[1];
      int r = query[2];
      ll sum = 0;
      for(int k = 0; k < LOGMAX; ++k){
        sum += (1ll << k) * tree[k].get(1, l, r);
      }
      res.push_back(sum);
    }
    else{
      int l = query[1];
      int r = query[2];
      int x = query[3];
      for(int k = 0; k < LOGMAX; ++k){
        if(1 & (x >> k)){
          tree[k].set(1, l, r);
        }
      }
    }
  }
  return res;
}

vector<ll> brute(int n, vector<vector<int>> &queries) {
  vector<int> arr(n + 1);
  vector<ll> res;
  for(vector<int> &query: queries){
    int type = query[0];
    if(type == 1){
      int l = query[1];
      int r = query[2];
      ll sum = 0;
      for(int i = l; i <= r; ++i){
        sum += arr[i];
      }
      res.push_back(sum);
    }
    else{
      int l = query[1];
      int r = query[2];
      int x = query[3];
      for(int i = l; i <= r; ++i){
        arr[i] = arr[i] ^ x;
      }
    }
  }
  return res;
}
 
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0); cout.tie(0);

  int n, q;
  cin >> n >> q;
  vector<vector<int>> queries(q);
  for(int i = 0; i < q; ++i){
    int type, l, r;
    cin >> type >> l >> r;
    queries[i] = {type, l, r};
    if(type == 2){
      int x;
      cin >> x;
      queries[i].push_back(x);
    }
  }

  for(ll elem: brute(n, queries)){
    cout << elem << '\n';
  }

  assert(solve(n, queries) == brute(n, queries));

  
  return 0;
}