description: You are given two integers n and m. Find the \operatorname{MEX} of the sequence n ⊕ 0, n ⊕ 1, …, n ⊕ m. Here, ⊕ is the [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
  
  \operatorname{MEX} of the sequence of non-negative integers is the smallest non-negative integer that doesn't appear in this sequence. For example, \operatorname{MEX}(0, 1, 2, 4) = 3, and \operatorname{MEX}(1, 2021) = 0. 
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 30 000) — the number of test cases.
  
  The first and only line of each test case contains two integers n and m (0 ≤ n, m ≤ 10^9).
  
  Output
  
  For each test case, print a single integer — the answer to the problem.
  
  Example
  
  Input
  
  
  5
  3 5
  4 6
  3 2
  69 696
  123456 654321
  
  
  Output
  
  
  4
  3
  0
  640
  530866
  
  Note
  
  In the first test case, the sequence is 3 ⊕ 0, 3 ⊕ 1, 3 ⊕ 2, 3 ⊕ 3, 3 ⊕ 4, 3 ⊕ 5, or 3, 2, 1, 0, 7, 6. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 4.
  
  In the second test case, the sequence is 4 ⊕ 0, 4 ⊕ 1, 4 ⊕ 2, 4 ⊕ 3, 4 ⊕ 4, 4 ⊕ 5, 4 ⊕ 6, or 4, 5, 6, 7, 0, 1, 2. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 3.
  
  In the third test case, the sequence is 3 ⊕ 0, 3 ⊕ 1, 3 ⊕ 2, or 3, 2, 1. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 0.
solution: #include <bits/stdc++.h>
  using namespace std;
  const int mod = 1e9 + 7;
  long long power(long long x, long long y, long long m) {
    if (y == 0) return 1;
    long long p = power(x, y / 2, m) % m;
    p = (p * p) % m;
    return (y % 2 == 0) ? p : (x * p) % m;
  }
  long long gcd(long long a, long long b) {
    if (a == 0) return b;
    return gcd(b % a, a);
  }
  long long modInverse(long long a, long long m) {
    long long g = gcd(a, m);
    if (g != 1)
      cout << "Inverse doesn't exist";
    else {
      return power(a, m - 2, m);
    }
  }
  long long log_a_b(long long a, long long b) {
    return (a > b - 1) ? 1 + log_a_b(a / b, b) : 0;
  }
  int t;
  vector<vector<long long> > graph;
  vector<long long> tin, tout;
  vector<bool> vis;
  vector<vector<long long> > up;
  void dfs(int v, int p, int l) {
    vis[v] = true;
    tin[v] = ++t;
    up[v][0] = p;
    for (int(i) = (1); (i) < (l + 1); (i)++) {
      up[v][i] = up[v][i - 1] < 0 ? -1 : up[up[v][i - 1]][i - 1];
    }
    for (int x : graph[v])
      if (!vis[x]) dfs(x, v, l);
    tout[v] = ++t;
  }
  bool isan(int u, int v) { return tin[u] <= tin[v] && tout[u] >= tout[v]; }
  int lca(int u, int v, int l) {
    if (isan(u, v)) return u;
    if (isan(v, u)) return v;
    for (int(i) = (l); (i) >= (0); (i)--) {
      if (up[u][i] == -1) continue;
      if (!isan(up[u][i], v)) u = up[u][i];
    }
    return up[u][0];
  }
  bool sum(long long n, long long i, long long s, long long cnt,
           vector<long long> &a) {
    if (i >= n && cnt > 0) {
      if (s != 0)
        return 0;
      else
        return 1;
    }
    if (i >= n && cnt == 0) return 0;
    bool a1 = sum(n, i + 1, s, cnt, a);
    bool b = sum(n, i + 1, s - a[i], cnt + 1, a);
    bool c = sum(n, i + 1, s + a[i], cnt + 1, a);
    bool d = (a1 | b | c);
    return d;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.precision(15);
    long long T;
    cin >> T;
    while (T--) {
      long long n, q, k, m, u, v, w, x;
      string s, t;
      cin >> n >> m;
      long long ans = 0;
      m++;
      vector<long long> a(31), b(31);
      for (int(i) = (0); (i) < (31); (i)++) {
        a[i] = n % 2;
        b[i] = m % 2;
        n /= 2;
        m /= 2;
      }
      reverse((a).begin(), (a).end());
      reverse((b).begin(), (b).end());
      for (int(i) = (0); (i) < (31); (i)++) {
        if (a[i] && !b[i]) break;
        if (!a[i] && b[i]) ans = ans | (1 << (30 - i));
      }
      cout << ans << "\n";
    }
    return 0;
  }
  