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;
  template <typename... Args>
  void logger(string vars, Args&&... values) {
    cout << vars << " = ";
    string delim = "";
    (..., (cout << delim << values, delim = ", "));
    cout << endl;
  }
  vector<bool> getBinary(int n) {
    vector<bool> res(32, false);
    int in = 31;
    while (n) {
      if (n & 1) res[in] = true;
      in--;
      n /= 2;
    }
    return res;
  }
  long long getDec(vector<bool> arr) {
    long long res = 0;
    for (int i = 0; i < arr.size(); i++) {
      res *= 2;
      if (arr[i]) {
        res += 1;
      }
    }
    return res;
  }
  void print(vector<bool> arr) {
    for (int i : arr) {
      cout << i;
    }
    cout << endl;
  }
  void solve() {
    int n, m;
    scanf("%d%d", &n, &m);
    if (n > m) {
      cout << "0" << endl;
      return;
    }
    vector<bool> nbin = getBinary(n);
    vector<bool> mbin = getBinary(m);
    vector<bool> res(32, false);
    bool found = false;
    for (int i = 0; i < 32; i++) {
      res[i] = nbin[i] ^ mbin[i];
      if (res[i] && nbin[i]) {
        res[i] = 0;
        found = true;
        break;
      }
    }
    if (!found) {
      for (int i = 31; i >= 0; i--) {
        if (!mbin[i]) {
          res[i] = 1;
          break;
        } else {
          res[i] = 0;
        }
      }
    }
    long long ans = getDec(res);
    cout << ans << endl;
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  