#include <bits/stdc++.h>
using namespace std;
int solve(int n, int m, int k) {
  if (n > m) return 0;
  if (n == 0) return min(m + 1, (1 << (k + 1)));
  if ((1 << k) & n) {
    int c = solve(n - (1 << k), m - (1 << k), k - 1);
    if (c == (1 << k))
      return (1 << (k + 1));
    else
      return c;
  } else {
    int z = solve(n, m - (1 << k), k - 1);
    if (z == (1 << k))
      return 2 * z;
    else {
      int c = solve(n, m, k - 1);
      if (c == (1 << k))
        return (1 << k) + z;
      else
        return c;
    }
  }
}
void solve() {
  int n, m;
  cin >> n >> m;
  cout << solve(n, m, 29) << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

### 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 solve(int n, int m, int k) {
  if (n > m) return 0;
  if (n == 0) return min(m + 1, (1 << (k + 1)));
  if ((1 << k) & n) {
    int c = solve(n - (1 << k), m - (1 << k), k - 1);
    if (c == (1 << k))
      return (1 << (k + 1));
    else
      return c;
  } else {
    int z = solve(n, m - (1 << k), k - 1);
    if (z == (1 << k))
      return 2 * z;
    else {
      int c = solve(n, m, k - 1);
      if (c == (1 << k))
        return (1 << k) + z;
      else
        return c;
    }
  }
}
void solve() {
  int n, m;
  cin >> n >> m;
  cout << solve(n, m, 29) << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

```