#include <bits/stdc++.h>
using namespace std;
vector<int> v[200005];
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    m++;
    string sn, sm;
    while (n > 0) {
      if (n % 2 == 1)
        sn = sn + '1';
      else
        sn = sn + '0';
      n = n / 2;
    }
    reverse(sn.begin(), sn.end());
    while (m > 0) {
      if (m % 2 == 1)
        sm = sm + '1';
      else
        sm = sm + '0';
      m = m / 2;
    }
    reverse(sm.begin(), sm.end());
    while (sn.size() < sm.size()) sn = '0' + sn;
    while (sm.size() < sn.size()) sm = '0' + sm;
    string ans;
    int tr = 0;
    for (int j = 0; j < sm.size(); j++) {
      if (sm[j] == '1' && sn[j] == '0' && tr == 0)
        ans = ans + '1';
      else
        ans = ans + '0';
      if (sn[j] == '1' && sm[j] == '0') tr = 1;
    }
    long long num = 0;
    long long k = 1;
    for (int j = ans.size() - 1; j >= 0; j--) {
      if (ans[j] == '1') num = num + k;
      k = k * 2;
    }
    cout << num << endl;
  }
}

### Prompt

Your task is to create 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;
vector<int> v[200005];
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    m++;
    string sn, sm;
    while (n > 0) {
      if (n % 2 == 1)
        sn = sn + '1';
      else
        sn = sn + '0';
      n = n / 2;
    }
    reverse(sn.begin(), sn.end());
    while (m > 0) {
      if (m % 2 == 1)
        sm = sm + '1';
      else
        sm = sm + '0';
      m = m / 2;
    }
    reverse(sm.begin(), sm.end());
    while (sn.size() < sm.size()) sn = '0' + sn;
    while (sm.size() < sn.size()) sm = '0' + sm;
    string ans;
    int tr = 0;
    for (int j = 0; j < sm.size(); j++) {
      if (sm[j] == '1' && sn[j] == '0' && tr == 0)
        ans = ans + '1';
      else
        ans = ans + '0';
      if (sn[j] == '1' && sm[j] == '0') tr = 1;
    }
    long long num = 0;
    long long k = 1;
    for (int j = ans.size() - 1; j >= 0; j--) {
      if (ans[j] == '1') num = num + k;
      k = k * 2;
    }
    cout << num << endl;
  }
}

```