#include <bits/stdc++.h>
using namespace std;
void decToBinary(int n, vector<int> &v) {
  for (int i = 31; i >= 0; i--) {
    int k = n >> i;
    if (k & 1)
      v.push_back(1);
    else
      v.push_back(0);
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int i, a, b, x = 0;
    cin >> a >> b;
    vector<int> v1, v2;
    decToBinary(a, v1);
    decToBinary(b, v2);
    for (i = 0; i < v1.size(); i++) {
      if (v1[i] == 1 && v2[i] == 1) x += pow(2, 31 - i);
    }
    int val = (a ^ x) + (b ^ x);
    cout << val << "\n";
  }
}

### Prompt

Please provide a cpp coded solution 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;
void decToBinary(int n, vector<int> &v) {
  for (int i = 31; i >= 0; i--) {
    int k = n >> i;
    if (k & 1)
      v.push_back(1);
    else
      v.push_back(0);
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int i, a, b, x = 0;
    cin >> a >> b;
    vector<int> v1, v2;
    decToBinary(a, v1);
    decToBinary(b, v2);
    for (i = 0; i < v1.size(); i++) {
      if (v1[i] == 1 && v2[i] == 1) x += pow(2, 31 - i);
    }
    int val = (a ^ x) + (b ^ x);
    cout << val << "\n";
  }
}

```