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;
  int calc(int n, int m) {
    int ans = 0;
    for (int k = 29; k >= 0; k--) {
      if (n > m) break;
      if (m + 1 == (1 << (k + 1))) {
        ans ^= (1 << (k + 1));
        break;
      }
      int nbit = ((n >> k) & 1);
      int mbit = ((m >> k) & 1);
      if (nbit == 0) {
        if (mbit == 1) {
          ans ^= (1 << k);
          m ^= (1 << k);
        }
      } else {
        if (mbit == 1) {
          n ^= (1 << k);
          m ^= (1 << k);
        } else {
          cerr << n << ' ' << m << endl;
          assert(false);
        }
      }
    }
    if (n == 0 && m == 0) ans++;
    return ans;
  }
  int calc_naive(int n, int m) {
    if (n > m) return 0;
    vector<int> b(2 * m + 2);
    for (int i = 0; i <= m; i++) {
      assert((n ^ i) < b.size());
      b[n ^ i] = 1;
    }
    for (int i = 0; i < b.size(); i++)
      if (b[i] == 0) return i;
    assert(false);
  }
  void solve() {
    int n, m;
    cin >> n >> m;
    int ans = calc(n, m);
    cout << ans << '\n';
  }
  void test_solve() {
    for (int n = 0; n <= 32; n++) {
      for (int m = 0; m <= 128; m++) {
        int r1 = calc(n, m);
        int r2 = calc_naive(n, m);
        if (r1 != r2) {
          cout << n << ' ' << m << ' ' << r1 << ' ' << r2 << endl;
          assert(false);
        }
      }
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int T;
    cin >> T;
    while (T--) {
      solve();
    }
  }
  