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 n, m;
vector<long long> holes;
long long _2[64];
long long masked(long long mask) {
  long long result = 0;
  for (int i = 0; i < 30; i++) {
    if (mask & _2[i]) {
      result = result + (_2[holes[i]]);
    }
  }
  return result;
}
int main() {
  _2[0] = 1;
  for (int i = 1; i <= 62; i++) {
    _2[i] = _2[i - 1] * 2;
  }
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> m;
    if (n > m) {
      cout << "0\n";
      continue;
    }
    holes.clear();
    for (int i = 0; i < 61; i++) {
      if (_2[i] & n)
        continue;
      else
        holes.push_back(i);
      if (holes.size() == 30) break;
    }
    long long l = 0;
    long long r = _2[30] + 5;
    long long ans = -1;
    while (l <= r) {
      long long mid = (l + r) / 2;
      if (masked(mid) + n > m) {
        ans = masked(mid);
        r = mid - 1;
      } else {
        l = mid + 1;
      }
    }
    cout << ans << endl;
  }
  return 0;
}
