#include <bits/stdc++.h>
using namespace std;
int main() {
  int T;
  cin >> T;
  while (T--) {
    int N, M;
    cin >> N >> M;
    if (M < N)
      cout << 0 << endl;
    else {
      int r = N, s = M + 1;
      int cnt = 0;
      while (r) {
        r >>= 1;
        s >>= 1;
        cnt++;
      }
      s <<= cnt;
      if (N > M - s)
        cout << s << endl;
      else {
        int add = 0;
        for (int i = cnt - 1; i >= 0; i--) {
          if ((N & (1 << i)) && !((M + 1) & (1 << i))) {
            break;
          } else {
            add += ((N & (1 << i)) ^ ((M + 1) & (1 << i)));
          }
        }
        cout << s + add << endl;
      }
    }
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

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 main() {
  int T;
  cin >> T;
  while (T--) {
    int N, M;
    cin >> N >> M;
    if (M < N)
      cout << 0 << endl;
    else {
      int r = N, s = M + 1;
      int cnt = 0;
      while (r) {
        r >>= 1;
        s >>= 1;
        cnt++;
      }
      s <<= cnt;
      if (N > M - s)
        cout << s << endl;
      else {
        int add = 0;
        for (int i = cnt - 1; i >= 0; i--) {
          if ((N & (1 << i)) && !((M + 1) & (1 << i))) {
            break;
          } else {
            add += ((N & (1 << i)) ^ ((M + 1) & (1 << i)));
          }
        }
        cout << s + add << endl;
      }
    }
  }
  return 0;
}

```