#include <bits/stdc++.h>
using std::bitset;
using std::ceil;
using std::cerr;
using std::cin;
using std::cout;
using std::deque;
using std::endl;
using std::floor;
using std::hash;
using std::make_pair;
using std::map;
using std::max;
using std::min;
using std::pair;
using std::priority_queue;
using std::queue;
using std::set;
using std::sort;
using std::stack;
using std::string;
using std::swap;
using std::unordered_map;
using std::vector;
const int MX = 31;
int n;
long int x, y;
inline void solve() {
  cin >> x >> y;
  bitset<MX> a(x);
  bitset<MX> b(y);
  for (int i = 0; i <= 30; i++) {
    if (a[i] == 1 && b[i] == 1) {
      a[i] = 0;
      b[i] = 0;
    }
  }
  long int k = a.to_ulong() + b.to_ulong();
  cout << k << endl;
}
int main() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t-- > 0) solve();
  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 std::bitset;
using std::ceil;
using std::cerr;
using std::cin;
using std::cout;
using std::deque;
using std::endl;
using std::floor;
using std::hash;
using std::make_pair;
using std::map;
using std::max;
using std::min;
using std::pair;
using std::priority_queue;
using std::queue;
using std::set;
using std::sort;
using std::stack;
using std::string;
using std::swap;
using std::unordered_map;
using std::vector;
const int MX = 31;
int n;
long int x, y;
inline void solve() {
  cin >> x >> y;
  bitset<MX> a(x);
  bitset<MX> b(y);
  for (int i = 0; i <= 30; i++) {
    if (a[i] == 1 && b[i] == 1) {
      a[i] = 0;
      b[i] = 0;
    }
  }
  long int k = a.to_ulong() + b.to_ulong();
  cout << k << endl;
}
int main() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t-- > 0) solve();
  return 0;
}

```