#include <bits/stdc++.h>
using namespace std;
struct P {
  int x, y;
  bool operator<(const P &p) {
    if (x != p.x)
      return x < p.x;
    else
      return y < p.y;
  }
};
void subsetGenerate(int n) {
  for (int b = 0; b < (1 << n); b++) {
    vector<int> subset;
    for (int i = 0; i < n; i++) {
      if (b & (1 << i)) subset.push_back(i);
    }
  }
}
void permutationGenerate(int n) {
  vector<int> permutation;
  for (int i = 0; i < n; i++) {
    permutation.push_back(i);
  }
  do {
  } while (next_permutation(permutation.begin(), permutation.end()));
}
bool customSort(int a, int b) { return a < b; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int a, b;
    cin >> a >> b;
    cout << (a ^ b) << 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;
struct P {
  int x, y;
  bool operator<(const P &p) {
    if (x != p.x)
      return x < p.x;
    else
      return y < p.y;
  }
};
void subsetGenerate(int n) {
  for (int b = 0; b < (1 << n); b++) {
    vector<int> subset;
    for (int i = 0; i < n; i++) {
      if (b & (1 << i)) subset.push_back(i);
    }
  }
}
void permutationGenerate(int n) {
  vector<int> permutation;
  for (int i = 0; i < n; i++) {
    permutation.push_back(i);
  }
  do {
  } while (next_permutation(permutation.begin(), permutation.end()));
}
bool customSort(int a, int b) { return a < b; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int a, b;
    cin >> a >> b;
    cout << (a ^ b) << endl;
  }
  return 0;
}

```