#include <bits/stdc++.h>
using namespace std;
long long T, n, m, x, y, ans;
signed main() {
  cin >> T;
  while (T--) {
    n = m = x = y = ans = 0;
    scanf("%lld%lld", &n, &m);
    y = m + 1;
    x = n;
    if (m < n) {
      puts("0");
      continue;
    }
    long long a[10005], b[10005];
    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));
    long long len1 = 0;
    long long len2 = 0;
    while (y) {
      a[++len1] = x & 1;
      b[++len2] = y & 1;
      x >>= 1;
      y >>= 1;
    }
    a[++len1] = 0;
    b[++len2] = 0;
    for (long long i = len2 - 1; i >= 0; i--) {
      if (a[i] == 0 && b[i] == 1) {
        ans += pow(2, i);
        a[i] == 1;
      } else if (a[i] == 1 && b[i] == 0)
        break;
      else
        continue;
    }
    printf("%lld\n", ans / 2);
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
long long T, n, m, x, y, ans;
signed main() {
  cin >> T;
  while (T--) {
    n = m = x = y = ans = 0;
    scanf("%lld%lld", &n, &m);
    y = m + 1;
    x = n;
    if (m < n) {
      puts("0");
      continue;
    }
    long long a[10005], b[10005];
    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));
    long long len1 = 0;
    long long len2 = 0;
    while (y) {
      a[++len1] = x & 1;
      b[++len2] = y & 1;
      x >>= 1;
      y >>= 1;
    }
    a[++len1] = 0;
    b[++len2] = 0;
    for (long long i = len2 - 1; i >= 0; i--) {
      if (a[i] == 0 && b[i] == 1) {
        ans += pow(2, i);
        a[i] == 1;
      } else if (a[i] == 1 && b[i] == 0)
        break;
      else
        continue;
    }
    printf("%lld\n", ans / 2);
  }
  return 0;
}

```