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;
class Solution {
 public:
  long long solve(int n, int m) {
    ++m;
    int h = 0;
    while (1LL << (h + 1) <= max(m, n)) {
      ++h;
    }
    long long mask = 0;
    long long prefix = 0;
    vector<pair<long long, long long>> ranges;
    for (int i = h; i >= 0; --i) {
      mask |= 1LL << i;
      if ((m >> i) & 1) {
        long long P = (n & mask) ^ prefix;
        long long mn = P;
        long long mx = P + (1LL << i) - 1;
        ranges.emplace_back(mn, mx);
        prefix |= 1LL << i;
      }
    }
    sort(ranges.begin(), ranges.end());
    int res = 0;
    for (auto& p : ranges) {
      long long l = p.first;
      long long r = p.second;
      if (l != res) {
        break;
      }
      res = r + 1;
    }
    return res;
  }
};
int main(int argc, char** argv) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  Solution sol;
  int t;
  cin >> t;
  while (t-- > 0) {
    int n, m;
    cin >> n >> m;
    cout << sol.solve(n, m) << '\n';
  }
  return 0;
}
