#include <bits/stdc++.h>
using namespace std;
long long expo(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  long long z = expo(a, b / 2);
  z *= z;
  z %= 1000000007;
  if (b % 2 != 0) {
    z *= a;
    z %= 1000000007;
  }
  return z;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
void solve() {
  long long n, m;
  cin >> n;
  cin >> m;
  if (m < n) {
    cout << "0\n";
    return;
  }
  vector<int> v1(32, 0);
  vector<int> v2(32, 0);
  int i = 0;
  while (n > 0) {
    v1[i] = n % 2;
    n /= 2;
    i++;
  }
  int j = 0;
  bool f = true;
  while (m > 0) {
    v2[j] = m % 2;
    if (v2[j] == 0) {
      f = false;
    }
    m /= 2;
    j++;
  }
  long long ans = 0;
  if (f) {
    ans = (1ll << j);
    cout << ans << "\n";
    return;
  }
  f = true;
  for (int j1 = 31; j1 >= 0; j1--) {
    if (v1[j1] == 1 && v2[j1] == 0) {
      f = false;
      break;
    }
  }
  if (f) {
    long long ansf = 100000000000000000;
    ans = 0;
    bool first = false;
    for (int j1 = 31; j1 >= 0; j1--) {
      if (v1[j1] == 0 && v2[j1] == 1) {
        first = true;
        ans += (1ll << j1);
      } else if (v1[j1] == 0 && v2[j1] == 0) {
        long long ans2 = (1ll << j1);
        ans2 += ans;
        ansf = min(ansf, ans2);
      }
    }
    cout << ansf << "\n";
    return;
  }
  long long ansf = 100000000000000000;
  for (int j1 = 31; j1 >= 0; j1--) {
    if (v1[j1] == 0 && v2[j1] == 1) {
      ans += (1ll << j1);
    } else if (v1[j1] == 0 && v2[j1] == 0) {
      long long ans2 = (1ll << j1);
      ans2 += ans;
      ansf = min(ansf, ans2);
    } else if (v1[j1] == 1 && v2[j1] == 0) {
      break;
    }
  }
  ansf = min(ans, ansf);
  cout << ansf << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
long long expo(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  long long z = expo(a, b / 2);
  z *= z;
  z %= 1000000007;
  if (b % 2 != 0) {
    z *= a;
    z %= 1000000007;
  }
  return z;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
void solve() {
  long long n, m;
  cin >> n;
  cin >> m;
  if (m < n) {
    cout << "0\n";
    return;
  }
  vector<int> v1(32, 0);
  vector<int> v2(32, 0);
  int i = 0;
  while (n > 0) {
    v1[i] = n % 2;
    n /= 2;
    i++;
  }
  int j = 0;
  bool f = true;
  while (m > 0) {
    v2[j] = m % 2;
    if (v2[j] == 0) {
      f = false;
    }
    m /= 2;
    j++;
  }
  long long ans = 0;
  if (f) {
    ans = (1ll << j);
    cout << ans << "\n";
    return;
  }
  f = true;
  for (int j1 = 31; j1 >= 0; j1--) {
    if (v1[j1] == 1 && v2[j1] == 0) {
      f = false;
      break;
    }
  }
  if (f) {
    long long ansf = 100000000000000000;
    ans = 0;
    bool first = false;
    for (int j1 = 31; j1 >= 0; j1--) {
      if (v1[j1] == 0 && v2[j1] == 1) {
        first = true;
        ans += (1ll << j1);
      } else if (v1[j1] == 0 && v2[j1] == 0) {
        long long ans2 = (1ll << j1);
        ans2 += ans;
        ansf = min(ansf, ans2);
      }
    }
    cout << ansf << "\n";
    return;
  }
  long long ansf = 100000000000000000;
  for (int j1 = 31; j1 >= 0; j1--) {
    if (v1[j1] == 0 && v2[j1] == 1) {
      ans += (1ll << j1);
    } else if (v1[j1] == 0 && v2[j1] == 0) {
      long long ans2 = (1ll << j1);
      ans2 += ans;
      ansf = min(ansf, ans2);
    } else if (v1[j1] == 1 && v2[j1] == 0) {
      break;
    }
  }
  ansf = min(ans, ansf);
  cout << ansf << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```