#include <bits/stdc++.h>
using namespace std;
int n, m, t;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> t;
  while (t--) {
    cin >> n >> m;
    if (n > m) {
      cout << 0 << "\n";
      continue;
    }
    int sol = n;
    for (int i = 31; i > -1; i--) {
      if (!((sol) & (1 << (i))) && ((m) & (1 << (i)))) {
        sol += 1 << i;
      }
      if (sol > m) break;
    }
    if (sol == m) {
      for (int i = 0; i <= 31; i++) {
        if (!((sol) & (1 << (i)))) {
          sol += 1 << i;
          break;
        }
        if (!((n) & (1 << (i))) && ((sol) & (1 << (i)))) {
          sol -= 1 << i;
        }
      }
    }
    cout << (sol ^ n) << "\n";
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, t;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> t;
  while (t--) {
    cin >> n >> m;
    if (n > m) {
      cout << 0 << "\n";
      continue;
    }
    int sol = n;
    for (int i = 31; i > -1; i--) {
      if (!((sol) & (1 << (i))) && ((m) & (1 << (i)))) {
        sol += 1 << i;
      }
      if (sol > m) break;
    }
    if (sol == m) {
      for (int i = 0; i <= 31; i++) {
        if (!((sol) & (1 << (i)))) {
          sol += 1 << i;
          break;
        }
        if (!((n) & (1 << (i))) && ((sol) & (1 << (i)))) {
          sol -= 1 << i;
        }
      }
    }
    cout << (sol ^ n) << "\n";
  }
}

```