#include <bits/stdc++.h>
using namespace std;
int n, m;
vector<pair<int, int>> vc;
void fun(int l, int r, int x) {
  if (r <= m) {
    int _l = (l ^ n) & (~x);
    int _r = _l + x;
    vc.push_back({_l, _r});
  } else {
    int mid = (l + r) >> 1;
    fun(l, mid, x >> 1);
    if (m > mid) fun(mid + 1, r, x >> 1);
  }
}
void solve() {
  scanf("%d %d", &n, &m);
  vc.clear();
  int x = 1;
  while (x <= m) x <<= 1;
  fun(0, x - 1, x - 1);
  sort(vc.begin(), vc.end());
  int ans = 0;
  for (auto &it : vc) {
    if (it.first != ans) {
      break;
    } else {
      ans = it.second + 1;
    }
  }
  printf("%d\n", ans);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
  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;
int n, m;
vector<pair<int, int>> vc;
void fun(int l, int r, int x) {
  if (r <= m) {
    int _l = (l ^ n) & (~x);
    int _r = _l + x;
    vc.push_back({_l, _r});
  } else {
    int mid = (l + r) >> 1;
    fun(l, mid, x >> 1);
    if (m > mid) fun(mid + 1, r, x >> 1);
  }
}
void solve() {
  scanf("%d %d", &n, &m);
  vc.clear();
  int x = 1;
  while (x <= m) x <<= 1;
  fun(0, x - 1, x - 1);
  sort(vc.begin(), vc.end());
  int ans = 0;
  for (auto &it : vc) {
    if (it.first != ans) {
      break;
    } else {
      ans = it.second + 1;
    }
  }
  printf("%d\n", ans);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}

```