#include <bits/stdc++.h>
using namespace std;
long long int mod(long long int x) {
  return ((x % 1000000007 + 1000000007) % 1000000007);
}
long long int add(long long int a, long long int b) {
  return mod(mod(a) + mod(b));
}
long long int mul(long long int a, long long int b) {
  return mod(mod(a) * mod(b));
}
bool cmp(const int &a, const int &b) { return a > b; }
bool comp2(pair<long long int, long long int> a,
           pair<long long int, long long int> b) {
  if (b.first > a.first) {
    return true;
  }
  if (b.first == a.first) {
    if (b.second > a.second)
      return true;
    else
      return false;
  } else
    return false;
}
bool isPowerOfTwo(int x) { return (x && !(x & (x - 1))); }
int count_one(int n) {
  long long int count = 0;
  while (n) {
    n = n & (n - 1);
    count++;
  }
  return count;
}
bool check_set(int N, int i) {
  if (N & (1 << i))
    return true;
  else
    return false;
}
void possibleSubsets(char A[], int N) {
  for (int i = 0; i < (1 << N); ++i) {
    for (int j = 0; j < N; ++j)
      if (i & (1 << j)) cout << A[j];
    cout << endl;
  }
}
vector<long long int> v[100005];
long long int vis[100005];
void dfs(long long int node) {
  vis[node] = 1;
  for (auto child : v[node]) {
    if (vis[child] == 0) {
      dfs(child);
    }
  }
}
long long int solve() {
  long long int n;
  cin >> n;
  ;
  long long int m;
  cin >> m;
  ;
  long long int ma = 0;
  for (int i = 32; i >= 0 && n < m + 1; i--) {
    if ((n >> i & 1) == ((m + 1) >> i & 1)) continue;
    if ((m + 1) >> i & 1) {
      ma |= 1 << i;
      n |= 1 << i;
    }
  }
  return ma;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  long long int t;
  cin >> t;
  ;
  while (t--) {
    cout << solve() << endl;
  }
  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 int mod(long long int x) {
  return ((x % 1000000007 + 1000000007) % 1000000007);
}
long long int add(long long int a, long long int b) {
  return mod(mod(a) + mod(b));
}
long long int mul(long long int a, long long int b) {
  return mod(mod(a) * mod(b));
}
bool cmp(const int &a, const int &b) { return a > b; }
bool comp2(pair<long long int, long long int> a,
           pair<long long int, long long int> b) {
  if (b.first > a.first) {
    return true;
  }
  if (b.first == a.first) {
    if (b.second > a.second)
      return true;
    else
      return false;
  } else
    return false;
}
bool isPowerOfTwo(int x) { return (x && !(x & (x - 1))); }
int count_one(int n) {
  long long int count = 0;
  while (n) {
    n = n & (n - 1);
    count++;
  }
  return count;
}
bool check_set(int N, int i) {
  if (N & (1 << i))
    return true;
  else
    return false;
}
void possibleSubsets(char A[], int N) {
  for (int i = 0; i < (1 << N); ++i) {
    for (int j = 0; j < N; ++j)
      if (i & (1 << j)) cout << A[j];
    cout << endl;
  }
}
vector<long long int> v[100005];
long long int vis[100005];
void dfs(long long int node) {
  vis[node] = 1;
  for (auto child : v[node]) {
    if (vis[child] == 0) {
      dfs(child);
    }
  }
}
long long int solve() {
  long long int n;
  cin >> n;
  ;
  long long int m;
  cin >> m;
  ;
  long long int ma = 0;
  for (int i = 32; i >= 0 && n < m + 1; i--) {
    if ((n >> i & 1) == ((m + 1) >> i & 1)) continue;
    if ((m + 1) >> i & 1) {
      ma |= 1 << i;
      n |= 1 << i;
    }
  }
  return ma;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  long long int t;
  cin >> t;
  ;
  while (t--) {
    cout << solve() << endl;
  }
  return 0;
}

```