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;
  void read(long long &x) {
    char ch = getchar();
    long long f = 0;
    x = 0;
    while (!isdigit(ch)) {
      if (ch == '-') f = 1;
      ch = getchar();
    }
    while (isdigit(ch)) x = x * 10 + ch - 48, ch = getchar();
    if (f) x = -x;
  }
  long long n, m, pw[35];
  signed main() {
    pw[0] = 1;
    for (long long i = 1; i <= 30; ++i) pw[i] = pw[i - 1] * 2;
    long long T;
    read(T);
    while (T--) {
      read(n), read(m);
      if (n > m) {
        puts("0");
        continue;
      }
      if (n == 0) {
        printf("%lld\n", m + 1);
        continue;
      }
      if (m == 0) {
        puts("0");
        continue;
      }
      long long p = n ^ m, mxa = 0, mxb = 0;
      for (long long i = 0; i <= 30; ++i) {
        if (n & pw[i]) mxa = i;
        if (m & pw[i]) mxb = i;
      }
      long long tag = 0;
      for (long long i = max(mxa, mxb); i >= 0; --i)
        if (p & pw[i]) {
          if ((n ^ (p ^ pw[i])) > m) p ^= pw[i], tag = 1;
        }
      if (!tag) {
        for (long long i = 0; i <= 30; ++i) {
          if (p & pw[i])
            p ^= pw[i];
          else if (((p ^ pw[i]) ^ n) > m) {
            p ^= pw[i];
            break;
          }
        }
      }
      printf("%lld\n", p);
    }
  }
  