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 dbg() { std::cout << "  #\n"; }
  template <typename T, typename... Args>
  void dbg(T a, Args... args) {
    std::cout << a << ' ';
    dbg(args...);
  }
  const int maxn = 2e5 + 10;
  const int MAX = (1 << 30);
  const int inf = 0x3f3f3f3f;
  const int mod = 1e9 + 7;
  int n, m;
  int a[maxn];
  int b[maxn];
  int getint() {
    char ch;
    while (!isdigit(ch = getchar()))
      ;
    int x = ch - 48;
    while (isdigit(ch = getchar())) x = x * 10 + ch - 48;
    return x;
  }
  vector<pair<int, int> > allseg;
  void divide(int l, int r, int L, int R, int val) {
    if (L > r || R < l) return;
    if (L <= l && r <= R) {
      l ^= val, r ^= val;
      allseg.push_back({l & r, l | r});
      return;
    }
    int mid = (l + r) >> 1;
    divide(l, mid, L, R, val);
    divide(mid + 1, r, L, R, val);
  }
  void solve() {
    n = getint();
    m = getint();
    allseg.clear();
    divide(0, MAX - 1, 0, m, n);
    sort((allseg).begin(), (allseg).end());
    int ans = 0;
    for (int i = 0; i < allseg.size(); i++) {
      pair<int, int> cur = allseg[i];
      if (cur.first <= ans)
        ans = max(ans, cur.second + 1);
      else {
        cout << (ans) << endl;
        return;
      }
    }
    cout << (ans) << endl;
    return;
  }
  signed main() {
    int TestCase = 1;
    cin >> TestCase;
    while (TestCase--) {
      solve();
    }
    char EndFile = getchar();
    EndFile = getchar();
    return 0;
  }
  