#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int mx = (1 << 30) - 1;
int n, m;
struct node {
  vector<pair<int, int> > p;
  void cal(int a, int b) {
    int res = a ^ b;
    for (int i = 0; i < 30; i++) {
      int t = res ^ (1 << i) ^ (res & ((1 << i) - 1));
      if ((t ^ a) > b) p.emplace_back(t, t + (1 << i) - 1);
    }
  }
  void get(int first, int l, int r) {
    cal(first, r);
    cal(mx ^ first, mx ^ l);
  }
};
void solve() {
  node temp;
  scanf("%d%d", &n, &m);
  temp.get(n, 0, m);
  sort(temp.p.begin(), temp.p.end());
  printf("%d\n", temp.p[0].first);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
}

### Prompt

Develop a solution in Cpp 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;
const int N = 1e5 + 5;
const int mx = (1 << 30) - 1;
int n, m;
struct node {
  vector<pair<int, int> > p;
  void cal(int a, int b) {
    int res = a ^ b;
    for (int i = 0; i < 30; i++) {
      int t = res ^ (1 << i) ^ (res & ((1 << i) - 1));
      if ((t ^ a) > b) p.emplace_back(t, t + (1 << i) - 1);
    }
  }
  void get(int first, int l, int r) {
    cal(first, r);
    cal(mx ^ first, mx ^ l);
  }
};
void solve() {
  node temp;
  scanf("%d%d", &n, &m);
  temp.get(n, 0, m);
  sort(temp.p.begin(), temp.p.end());
  printf("%d\n", temp.p[0].first);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
}

```