#include <bits/stdc++.h>
template <class C>
C& mini(C& a, C b) {
  if (b < a) a = b;
  return a;
}
template <class C>
C& maxi(C& a, C b) {
  if (a < b) a = b;
  return a;
}
template <typename T>
void read(std::vector<T>& A) {
  for (T& x : A) std::cin >> x;
}
template <typename T>
void readv(std::vector<T>& A) {
  int N;
  std::cin >> N;
  A.resize(N);
  read(A);
}
using namespace std;
int getmask(int N) {
  int b = 1;
  while (b <= N) b <<= 1;
  return b - 1;
}
int solve(int N, int M) {
  if (N == 0) return M + 1;
  if (N > M) return 0;
  int m = getmask(N);
  int left = M & ~m, right = M & m;
  if (left == 0) {
    assert(right);
    if (((right + 1) & right) == 0) return M + 1;
    int b = (m + 1) >> 1;
    assert(N & b);
    assert(M & b);
    return solve(N ^ b, M ^ b);
  } else {
    if ((right & m) == m) return M + 1;
    return left ^ solve(N, M ^ left);
  }
}
int main(int argc, const char* argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(20);
  cout.setf(ios::fixed);
  int T;
  cin >> T;
  for (int t = 1; t <= T; t++) {
    int N, M;
    cin >> N >> M;
    cout << solve(N, M) << '\n';
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution 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>
template <class C>
C& mini(C& a, C b) {
  if (b < a) a = b;
  return a;
}
template <class C>
C& maxi(C& a, C b) {
  if (a < b) a = b;
  return a;
}
template <typename T>
void read(std::vector<T>& A) {
  for (T& x : A) std::cin >> x;
}
template <typename T>
void readv(std::vector<T>& A) {
  int N;
  std::cin >> N;
  A.resize(N);
  read(A);
}
using namespace std;
int getmask(int N) {
  int b = 1;
  while (b <= N) b <<= 1;
  return b - 1;
}
int solve(int N, int M) {
  if (N == 0) return M + 1;
  if (N > M) return 0;
  int m = getmask(N);
  int left = M & ~m, right = M & m;
  if (left == 0) {
    assert(right);
    if (((right + 1) & right) == 0) return M + 1;
    int b = (m + 1) >> 1;
    assert(N & b);
    assert(M & b);
    return solve(N ^ b, M ^ b);
  } else {
    if ((right & m) == m) return M + 1;
    return left ^ solve(N, M ^ left);
  }
}
int main(int argc, const char* argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(20);
  cout.setf(ios::fixed);
  int T;
  cin >> T;
  for (int t = 1; t <= T; t++) {
    int N, M;
    cin >> N >> M;
    cout << solve(N, M) << '\n';
  }
  return 0;
}

```