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("Ofast")
  #pragma GCC target("avx,avx2,fma")
  #pragma GCC optimization("O3")
  #pragma GCC optimization("unroll-loops")
  using namespace std;
  const int mod = 1e9 + 7;
  const int N = 1e6 + 10;
  const long long inf = 1e18;
  const int maxm = 1005;
  vector<vector<char>> matrix(maxm, vector<char>(maxm));
  struct event {
    int x, y, size;
  };
  int dx[] = {1, 0, -1, 0};
  int dy[] = {0, -1, 0, 1};
  struct myComp {
    constexpr bool operator()(pair<int, int> const& a,
                              pair<int, int> const& b) const noexcept {
      return a.first < b.first;
    }
  };
  class even {
   public:
    int a, b, c, ind;
  };
  long long poww(long long a, long long b, long long m) {
    if (b == 0) return 1ll;
    if (b % 2 == 1) return a * poww(a, b - 1, m) % m;
    long long rs = poww(a, b / 2, m);
    rs = rs * rs % m;
    return rs;
  }
  long long fact[N];
  long long ncr(long long n, long long r) {
    long long num = fact[n];
    long long deno = fact[n - r] * fact[r] % mod;
    deno = poww(deno, mod - 2, mod);
    num = num * deno;
    num = num % mod;
    return num;
  }
  vector<int> parent(N), st(N);
  void set_parent() {
    for (int i = 0; i < N; i++) parent[i] = i, st[i] = 1;
  }
  int find_parent(int v) {
    if (parent[v] == v)
      return v;
    else
      return parent[v] = find_parent(parent[v]);
  }
  int x = 1;
  void union_set(int a, int b) {
    int temp = a;
    if (a == b) return;
    a = find_parent(a);
    b = find_parent(b);
    if (a == b) {
      x = temp;
      return;
    }
    if (st[a] < st[b]) swap(a, b);
    st[a] += st[b];
    parent[b] = a;
  }
  const int maxx = 1e7 + 10;
  vector<long long> spf(maxx, -1);
  void sieve() {
    spf[1] = 1;
    for (long long i = 2; i * i < maxx; i++) {
      if (spf[i] == -1) {
        spf[i] = i;
        for (long long j = i * i; j < maxx; j += i) {
          if (spf[j] == -1) spf[j] = i;
        }
      }
    }
  }
  bool ok(int mid, int n, int m) {
    int res = 0;
    for (int i = 0; i < 32; i++) {
      if (mid & (1 << i)) {
        if (n & (1 << i)) {
        } else
          res += 1 << i;
      } else {
        if (n & (1 << i)) res += 1 << i;
      }
    }
    cout << mid << " " << res << " " << m << "\n";
    if (res <= m) return true;
    return false;
  }
  void solve() {
    long long n, m;
    cin >> n >> m;
    long long val = 0;
    long long ans = 1e10;
    for (long long i = 0; i < 32; i++)
      if (!(m >> i & 1ll)) {
        long long temp = 0;
        if (n >> i & 1ll) {
        } else
          temp += 1ll << i;
        for (long long j = i + 1ll; j < 32; j++) {
          if (m >> j & 1ll)
            if (!(n >> j & 1ll)) temp += 1ll << j;
        }
        ans = min(ans, temp);
      }
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) solve();
  }
  