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 main() {
    long long t;
    cin >> t;
    while (t--) {
      long long n, m;
      cin >> n >> m;
      if (n == 0 && m == 0)
        printf("1\n");
      else if (n > m)
        printf("0\n");
      else {
        long long a[30 + 1] = {0}, idx = 0;
        long long sum = 0, val = 1, cnt = 1;
        long long nn = n;
        while (nn != 0) {
          if (nn % 2 == 0) a[++idx] = val, cnt *= 2;
          nn /= 2;
          val *= 2;
        }
        a[0] = 1;
        while (val <= m * 2) a[++idx] = val, val *= 2, cnt *= 2;
        long long low = 0, high = cnt - 1;
        long long ans = 0;
        while (low <= high) {
          long long mid = (low + high) / 2;
          long long val = mid;
          long long bits[30 + 1] = {0}, i = 0;
          while (val != 0) {
            bits[++i] = val % 2;
            val /= 2;
          }
          long long sum = 0;
          for (long long j = 1; j <= idx; j++)
            if (bits[j] == 1) sum += a[j];
          if (sum + n <= m)
            low = mid + 1;
          else
            high = mid - 1, ans = sum;
        }
        printf("%lld\n", ans);
      }
    }
    return 0;
  }
  