#include <bits/stdc++.h>
using namespace std;
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    if (n == 0 && m == 0)
      printf("1\n");
    else if (n > m)
      printf("0\n");
    else {
      long long a[30 + 1] = {0}, idx = 0;
      long long sum = 0, val = 1, cnt = 1;
      long long nn = n;
      while (nn != 0) {
        if (nn % 2 == 0) a[++idx] = val, cnt *= 2;
        nn /= 2;
        val *= 2;
      }
      a[0] = 1;
      while (val <= m * 2) a[++idx] = val, val *= 2, cnt *= 2;
      long long low = 0, high = cnt - 1;
      long long ans = 0;
      while (low <= high) {
        long long mid = (low + high) / 2;
        long long val = mid;
        long long bits[30 + 1] = {0}, i = 0;
        while (val != 0) {
          bits[++i] = val % 2;
          val /= 2;
        }
        long long sum = 0;
        for (long long j = 1; j <= idx; j++)
          if (bits[j] == 1) sum += a[j];
        if (sum + n <= m)
          low = mid + 1;
        else
          high = mid - 1, ans = sum;
      }
      printf("%lld\n", ans);
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    if (n == 0 && m == 0)
      printf("1\n");
    else if (n > m)
      printf("0\n");
    else {
      long long a[30 + 1] = {0}, idx = 0;
      long long sum = 0, val = 1, cnt = 1;
      long long nn = n;
      while (nn != 0) {
        if (nn % 2 == 0) a[++idx] = val, cnt *= 2;
        nn /= 2;
        val *= 2;
      }
      a[0] = 1;
      while (val <= m * 2) a[++idx] = val, val *= 2, cnt *= 2;
      long long low = 0, high = cnt - 1;
      long long ans = 0;
      while (low <= high) {
        long long mid = (low + high) / 2;
        long long val = mid;
        long long bits[30 + 1] = {0}, i = 0;
        while (val != 0) {
          bits[++i] = val % 2;
          val /= 2;
        }
        long long sum = 0;
        for (long long j = 1; j <= idx; j++)
          if (bits[j] == 1) sum += a[j];
        if (sum + n <= m)
          low = mid + 1;
        else
          high = mid - 1, ans = sum;
      }
      printf("%lld\n", ans);
    }
  }
  return 0;
}

```