#include <bits/stdc++.h>
using namespace std;
int main() {
  long long t, n, k, a, b;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    cin >> a >> b;
    int x = 1, s = 0;
    while (x <= max(a, b)) {
      x = 2 * x;
    }
    x = x / 2;
    while (x >= 1) {
      int i = 0;
      if (a >= x) {
        a = a - x;
        i++;
      }
      if (b >= x) {
        b = b - x;
        i++;
      }
      if (i == 1) {
        s = s + x;
      }
      x = x / 2;
    }
    cout << s << endl;
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 main() {
  long long t, n, k, a, b;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    cin >> a >> b;
    int x = 1, s = 0;
    while (x <= max(a, b)) {
      x = 2 * x;
    }
    x = x / 2;
    while (x >= 1) {
      int i = 0;
      if (a >= x) {
        a = a - x;
        i++;
      }
      if (b >= x) {
        b = b - x;
        i++;
      }
      if (i == 1) {
        s = s + x;
      }
      x = x / 2;
    }
    cout << s << endl;
  }
  return 0;
}

```