#include <bits/stdc++.h>
using namespace std;
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << H;
  debug_out(T...);
}
const int INF = 0x3f3f3f3f, m197 = (int)1e9 + 7;
void s_solve() {
  int a, b;
  cin >> a >> b;
  long long res = 0;
  for (int i = 0; i <= 30; ++i) {
    if (((a >> i) & 1) && !((b >> i) & 1))
      res += 1 << i;
    else if (!((a >> i) & 1) && ((b >> i) & 1))
      res += 1 << i;
  }
  cout << res << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(10);
  int _T = 1, _C = 1;
  cin >> _T;
  while (_C++ <= _T) {
    s_solve();
  }
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

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 debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << H;
  debug_out(T...);
}
const int INF = 0x3f3f3f3f, m197 = (int)1e9 + 7;
void s_solve() {
  int a, b;
  cin >> a >> b;
  long long res = 0;
  for (int i = 0; i <= 30; ++i) {
    if (((a >> i) & 1) && !((b >> i) & 1))
      res += 1 << i;
    else if (!((a >> i) & 1) && ((b >> i) & 1))
      res += 1 << i;
  }
  cout << res << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(10);
  int _T = 1, _C = 1;
  cin >> _T;
  while (_C++ <= _T) {
    s_solve();
  }
  return 0;
}

```