#include <bits/stdc++.h>
using namespace std;
template <typename T>
using MinHeap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using MaxHeap = priority_queue<T>;
const long double PI = 3.14159265358979323846;
const long double E = 2.718281828459;
long long MOD = (1ll << 31) - 1;
long long M(long long n) { return (n % MOD + MOD) % MOD; }
long long pow2(long long n) { return 1ll << n; }
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
int main(int argn, char **argv) {
  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  int T = 1;
  cin >> T;
  for (long long(I) = (0); (I) <= ((T)-1); (I)++) {
    long long n, m;
    cin >> n >> m;
    long long ans = 0, cannot = 0;
    for (long long(bit) = (31); (bit) >= (0); (bit)--) {
      long long bt = 1ll << bit;
      long long lim = (ans ^ n) | (bt - 1);
      if (m >= lim) {
        ans += bt;
      }
    }
    if ((n ^ ans) <= m) ans++;
    ;
    cout << ans << endl;
  }
  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>
using namespace std;
template <typename T>
using MinHeap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using MaxHeap = priority_queue<T>;
const long double PI = 3.14159265358979323846;
const long double E = 2.718281828459;
long long MOD = (1ll << 31) - 1;
long long M(long long n) { return (n % MOD + MOD) % MOD; }
long long pow2(long long n) { return 1ll << n; }
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
int main(int argn, char **argv) {
  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  int T = 1;
  cin >> T;
  for (long long(I) = (0); (I) <= ((T)-1); (I)++) {
    long long n, m;
    cin >> n >> m;
    long long ans = 0, cannot = 0;
    for (long long(bit) = (31); (bit) >= (0); (bit)--) {
      long long bt = 1ll << bit;
      long long lim = (ans ^ n) | (bt - 1);
      if (m >= lim) {
        ans += bt;
      }
    }
    if ((n ^ ans) <= m) ans++;
    ;
    cout << ans << endl;
  }
  return 0;
}

```