#include <bits/stdc++.h>
using namespace std;
int dx[] = {0, 0, -1, 1, -1, 1, -1, 1};
int dy[] = {-1, 1, 0, 0, -1, 1, 1, -1};
const string alpha = "abcdefghijklmnopqrstuvwxyz";
long long const N = 16;
long long n, m, T, k;
int mate[N][N];
vector<vector<long long>> genSubsets(long long &n, vector<long long> arr) {
  vector<vector<long long>> res;
  vector<long long> temp;
  for (int i = 0; i < pow(2, n); i++) {
    temp.clear();
    for (int j = 0; j < n; j++) {
      if (i & (1 << j)) {
        temp.push_back(arr[j]);
      }
    }
    res.push_back(temp);
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> T;
  while (T--) {
    long long a, b;
    cin >> a >> b;
    cout << (a ^ b) << "\n";
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

In order to celebrate Twice's 5th anniversary, Tzuyu and Sana decided to play a game.

Tzuyu gave Sana two integers a and b and a really important quest.

In order to complete the quest, Sana has to output the smallest possible value of (a ⊕ x) + (b ⊕ x) for any given x, where ⊕ denotes the [bitwise XOR operation](http://en.wikipedia.org/wiki/Bitwise_operation#XOR). 

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^{4}). Description of the test cases follows.

The only line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^{9}).

Output

For each testcase, output the smallest possible value of the given expression.

Example

Input


6
6 12
4 9
59 832
28 14
4925 2912
1 1


Output


10
13
891
18
6237
0

Note

For the first test case Sana can choose x=4 and the value will be (6 ⊕ 4) + (12 ⊕ 4) = 2 + 8 = 10. It can be shown that this is the smallest possible value.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int dx[] = {0, 0, -1, 1, -1, 1, -1, 1};
int dy[] = {-1, 1, 0, 0, -1, 1, 1, -1};
const string alpha = "abcdefghijklmnopqrstuvwxyz";
long long const N = 16;
long long n, m, T, k;
int mate[N][N];
vector<vector<long long>> genSubsets(long long &n, vector<long long> arr) {
  vector<vector<long long>> res;
  vector<long long> temp;
  for (int i = 0; i < pow(2, n); i++) {
    temp.clear();
    for (int j = 0; j < n; j++) {
      if (i & (1 << j)) {
        temp.push_back(arr[j]);
      }
    }
    res.push_back(temp);
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> T;
  while (T--) {
    long long a, b;
    cin >> a >> b;
    cout << (a ^ b) << "\n";
  }
  return 0;
}

```