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>
  #pragma GCC optimize(2)
  using namespace std;
  const double pai = acos(-1.0);
  const int maxn = 1e6 + 10;
  const int mod = 998244353;
  const double eps = 1e-9;
  const int N = 5e3 + 10;
  inline int read() {
    int k = 0, f = 1;
    char c = getchar();
    while (!isdigit(c)) {
      if (c == '-') f = -1;
      c = getchar();
    }
    while (isdigit(c)) k = (k << 1) + (k << 3) + c - 48, c = getchar();
    return k * f;
  }
  long long n, m, mul[40];
  int t;
  int main() {
    mul[0] = 1;
    for (int i = 1; i <= 32; i++) mul[i] = mul[i - 1] * 2;
    scanf("%d", &t);
    while (t--) {
      scanf("%lld%lld", &n, &m);
      long long sum = 0, p = m + 1;
      for (int i = 32; i >= 0; i--) {
        long long x = n & (1ll << i), y = p & (1ll << i);
        if (x && !y) break;
        if (y && !x) sum += mul[i];
      }
      printf("%lld\n", sum);
    }
    return 0;
  }
  