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.
#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
void solve() {
  long long n, m;
  cin >> n >> m;
  if (n > m) {
    cout << "0\n";
    return;
  }
  vector<long long> pre(35, 0);
  long long res = 1;
  for (long long bit = 0; bit < 31; bit++) {
    if (!(n & res)) {
      if (bit == 0) {
        pre[bit] += res;
      } else {
        pre[bit] += pre[bit - 1];
        pre[bit] += res;
      }
    } else if (bit != 0)
      pre[bit] += pre[bit - 1];
    res <<= 1;
  }
  long long ans = m - n + 1;
  long long mul = 1 << 30;
  for (long long bit = 30; bit >= 0; bit--) {
    if (!(n & mul)) {
      if (bit == 0 && ans > 0) {
        ans -= mul;
      } else if (pre[bit - 1] < ans && ans > 0) {
        ans -= mul;
      }
    }
    mul >>= 1;
  }
  ans = abs(ans) + m + 1;
  ans = n ^ ans;
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}
