#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  int n, m;
  int i, j;
  int ans;
  vector<int> v1, v2;
  scanf("%d", &t);
  int flag = 0;
  while (t--) {
    v1.clear();
    v2.clear();
    flag = 0;
    scanf("%d%d", &n, &m);
    if (n == 0) {
      ans = m + 1;
      printf("%d\n", ans);
      continue;
    }
    if (n > m) {
      ans = 0;
      printf("%d\n", ans);
      continue;
    }
    int tmp = n;
    while (tmp > 0) {
      v1.push_back(tmp & 1);
      tmp = tmp >> 1;
    }
    tmp = m;
    while (tmp > 0) {
      v2.push_back(tmp & 1);
      tmp = tmp >> 1;
    }
    for (i = v1.size() - 1; i >= 0; i--) {
      if (v1[i] == v2[i]) continue;
      if (v1[i] > v2[i]) {
        flag = 1;
        break;
      }
    }
    int l;
    tmp = m + 1;
    j = 0;
    while (tmp > 0 && (tmp & 1) == 0) {
      tmp = tmp >> 1;
      j++;
    }
    ans = n ^ (m + 1);
    l = i + 1;
    ans = ans >> l;
    ans = ans << l;
    ans = ans >> j;
    ans = ans << j;
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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() {
  int t;
  int n, m;
  int i, j;
  int ans;
  vector<int> v1, v2;
  scanf("%d", &t);
  int flag = 0;
  while (t--) {
    v1.clear();
    v2.clear();
    flag = 0;
    scanf("%d%d", &n, &m);
    if (n == 0) {
      ans = m + 1;
      printf("%d\n", ans);
      continue;
    }
    if (n > m) {
      ans = 0;
      printf("%d\n", ans);
      continue;
    }
    int tmp = n;
    while (tmp > 0) {
      v1.push_back(tmp & 1);
      tmp = tmp >> 1;
    }
    tmp = m;
    while (tmp > 0) {
      v2.push_back(tmp & 1);
      tmp = tmp >> 1;
    }
    for (i = v1.size() - 1; i >= 0; i--) {
      if (v1[i] == v2[i]) continue;
      if (v1[i] > v2[i]) {
        flag = 1;
        break;
      }
    }
    int l;
    tmp = m + 1;
    j = 0;
    while (tmp > 0 && (tmp & 1) == 0) {
      tmp = tmp >> 1;
      j++;
    }
    ans = n ^ (m + 1);
    l = i + 1;
    ans = ans >> l;
    ans = ans << l;
    ans = ans >> j;
    ans = ans << j;
    printf("%d\n", ans);
  }
  return 0;
}

```