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;
  #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
  #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
  const long long mod = 1000 * 1000 * 1000 + 7;
  const long long inf = 1ll * 1000 * 1000 * 1000 * 1000 * 1000 * 1000 + 7;
  const long long mod2 = 998244353;
  const long long N = 1000 * 1000 + 10;
  const long double pi = 3.141592653589793;
  long long power(long long x, long long y, long long p = LLONG_MAX) {
    long long res = 1;
    x %= p;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  long long ncr(long long n, long long r, long long m) {
    if (r > n) return 0;
    long long a = 1, b = 1, i;
    for (i = 0; i < r; i++) {
      a = (a * n) % m;
      --n;
    }
    while (r) {
      b = (b * r) % m;
      --r;
    }
    return (a * power(b, m - 2, m)) % m;
  }
  void solve() {
    long long n, m;
    cin >> n >> m;
    bitset<64> a(n), b(0), c(m + 1);
    for (long long i = 63; i >= 0; i--) {
      if (a[i] == c[i]) continue;
      if (a[i] == 1 && c[i] == 0) {
        break;
      } else {
        b[i] = 1;
      }
    }
    long long ans = 0;
    for (long long i = 63; i >= 0; i--) ans += b[i] * pow(2, i);
    cout << ans << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  