#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
const int N = 2e5 + 5;
const long long int mod = 1e9 + 7;
void solve() {
  long long int n, m, i, ans = 0;
  cin >> n >> m;
  m++;
  long long int bit1, bit2;
  for (i = 31; i >= 0; i--) {
    bit1 = (n >> i) & 1;
    bit2 = (m >> i) & 1;
    if (bit1 == bit2) continue;
    if (bit1 && bit2 == 0) break;
    ans += (1 << i);
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Generate 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>
using namespace std;
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
const int N = 2e5 + 5;
const long long int mod = 1e9 + 7;
void solve() {
  long long int n, m, i, ans = 0;
  cin >> n >> m;
  m++;
  long long int bit1, bit2;
  for (i = 31; i >= 0; i--) {
    bit1 = (n >> i) & 1;
    bit2 = (m >> i) & 1;
    if (bit1 == bit2) continue;
    if (bit1 && bit2 == 0) break;
    ans += (1 << i);
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```