#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int dfs(int l, int r, int B, int a, int b, int x) {
  if (l == r || a > b) return l;
  int mid = (1ll * l + r) >> 1;
  if (x >> B & 1) {
    if (b == r && a <= mid + 1)
      return dfs(l + (1 << B), r, B - 1, a + (1 << B), min(r, b + (1 << B)), x);
    else
      return dfs(l, l + (1 << B) - 1, B - 1, max(l, a - (1 << B)), b - (1 << B),
                 x);
  } else {
    if (l == a && b >= mid)
      return dfs(l + (1 << B), r, B - 1, max(a, l + (1 << B)), b, x);
    else
      return dfs(l, l + (1 << B) - 1, B - 1, a, min(l + (1 << B) - 1, b), x);
  }
}
void work() {
  int n, m;
  scanf("%d%d", &n, &m);
  printf("%d\n", dfs(0, (1 << 30) - 1, 29, 0, m, n));
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) work();
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 int N = 1e5 + 5;
int dfs(int l, int r, int B, int a, int b, int x) {
  if (l == r || a > b) return l;
  int mid = (1ll * l + r) >> 1;
  if (x >> B & 1) {
    if (b == r && a <= mid + 1)
      return dfs(l + (1 << B), r, B - 1, a + (1 << B), min(r, b + (1 << B)), x);
    else
      return dfs(l, l + (1 << B) - 1, B - 1, max(l, a - (1 << B)), b - (1 << B),
                 x);
  } else {
    if (l == a && b >= mid)
      return dfs(l + (1 << B), r, B - 1, max(a, l + (1 << B)), b, x);
    else
      return dfs(l, l + (1 << B) - 1, B - 1, a, min(l + (1 << B) - 1, b), x);
  }
}
void work() {
  int n, m;
  scanf("%d%d", &n, &m);
  printf("%d\n", dfs(0, (1 << 30) - 1, 29, 0, m, n));
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) work();
}

```