#include <bits/stdc++.h>
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
#pragma GCC optimize("Ofast")
using namespace std;
long long int mod = 1000000007LL;
template <class A = long long int>
void Out(vector<A> a) {
  for (long long int i = (long long int)0; i < (long long int)a.size(); i++) {
    cout << a[i] << " ";
  }
  cout << endl;
}
template <class A = long long int>
void In(vector<A> &a, long long int N) {
  A cur;
  for (long long int i = (long long int)0; i < (long long int)N; i++) {
    cin >> cur;
    a.push_back(cur);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int T;
  cin >> T;
  for (long long int t = (long long int)0; t < (long long int)T; t++) {
    long long int N, M;
    cin >> N >> M;
    if (N > M) {
      cout << 0 << endl;
      continue;
    }
    M++;
    long long int D = (long long int)log2(4LL * M);
    long long int y = 0LL;
    for (long long int i = D - 1; i >= 0; i--) {
      long long int dN = N & (1LL << i);
      long long int dM = M & (1LL << i);
      if (dN != 0LL && dM == 0LL) {
        break;
      }
      if (dN == 0LL && dM != 0LL) {
        y += (1LL << i);
      }
    }
    cout << y << endl;
  }
  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>
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
#pragma GCC optimize("Ofast")
using namespace std;
long long int mod = 1000000007LL;
template <class A = long long int>
void Out(vector<A> a) {
  for (long long int i = (long long int)0; i < (long long int)a.size(); i++) {
    cout << a[i] << " ";
  }
  cout << endl;
}
template <class A = long long int>
void In(vector<A> &a, long long int N) {
  A cur;
  for (long long int i = (long long int)0; i < (long long int)N; i++) {
    cin >> cur;
    a.push_back(cur);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int T;
  cin >> T;
  for (long long int t = (long long int)0; t < (long long int)T; t++) {
    long long int N, M;
    cin >> N >> M;
    if (N > M) {
      cout << 0 << endl;
      continue;
    }
    M++;
    long long int D = (long long int)log2(4LL * M);
    long long int y = 0LL;
    for (long long int i = D - 1; i >= 0; i--) {
      long long int dN = N & (1LL << i);
      long long int dM = M & (1LL << i);
      if (dN != 0LL && dM == 0LL) {
        break;
      }
      if (dN == 0LL && dM != 0LL) {
        y += (1LL << i);
      }
    }
    cout << y << endl;
  }
  return 0;
}

```