#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int test = 1;
  cin >> test;
  while (test--) {
    long long int n, m, i, j;
    cin >> n >> m;
    long long int x = 0;
    if (m < n)
      cout << 0 << "\n";
    else {
      m++;
      long long int f = 0;
      for (i = 30; i > -1; i--) {
        long long int a, b;
        if (n & (1LL << i))
          a = 1;
        else
          a = 0;
        if (m & (1LL << i))
          b = 1;
        else
          b = 0;
        if (a == 0 && b == 0)
          ;
        else if (a == 1 && b == 0)
          x ^= (1LL << i), f = 1;
        else if (a == 1 && b == 1)
          x ^= (1LL << i);
        else {
          if (!f) x ^= (1LL << i);
        }
      }
      cout << (x ^ n) << "\n";
    }
  }
  cerr << "Time Taken:" << (float)clock() / CLOCKS_PER_SEC * 1000 << "ms"
       << "\n";
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const long long int MOD = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int test = 1;
  cin >> test;
  while (test--) {
    long long int n, m, i, j;
    cin >> n >> m;
    long long int x = 0;
    if (m < n)
      cout << 0 << "\n";
    else {
      m++;
      long long int f = 0;
      for (i = 30; i > -1; i--) {
        long long int a, b;
        if (n & (1LL << i))
          a = 1;
        else
          a = 0;
        if (m & (1LL << i))
          b = 1;
        else
          b = 0;
        if (a == 0 && b == 0)
          ;
        else if (a == 1 && b == 0)
          x ^= (1LL << i), f = 1;
        else if (a == 1 && b == 1)
          x ^= (1LL << i);
        else {
          if (!f) x ^= (1LL << i);
        }
      }
      cout << (x ^ n) << "\n";
    }
  }
  cerr << "Time Taken:" << (float)clock() / CLOCKS_PER_SEC * 1000 << "ms"
       << "\n";
  return 0;
}

```