#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n, m;
  scanf("%lld %lld", &n, &m);
  if (n > m) {
    printf("0\n");
    return;
  }
  m = m + 1;
  long long pow2 = 1;
  while (pow2 <= n) pow2 = pow2 * 2;
  pow2 = pow2 / 2;
  while (pow2 > 0) {
    if ((n & pow2) != 0 && (m & pow2) == 0) {
      pow2 = pow2 * 2;
      m = (m / pow2) * pow2;
      break;
    } else if ((n & pow2) != 0)
      m -= pow2;
    pow2 = pow2 / 2;
  }
  printf("%lld\n", m);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
}

### Prompt

Generate a Cpp solution to 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;
void solve() {
  long long n, m;
  scanf("%lld %lld", &n, &m);
  if (n > m) {
    printf("0\n");
    return;
  }
  m = m + 1;
  long long pow2 = 1;
  while (pow2 <= n) pow2 = pow2 * 2;
  pow2 = pow2 / 2;
  while (pow2 > 0) {
    if ((n & pow2) != 0 && (m & pow2) == 0) {
      pow2 = pow2 * 2;
      m = (m / pow2) * pow2;
      break;
    } else if ((n & pow2) != 0)
      m -= pow2;
    pow2 = pow2 / 2;
  }
  printf("%lld\n", m);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
}

```