#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
struct _ {
  ios_base::Init i;
  _() {
    cin.sync_with_stdio(0);
    cin.tie(0);
    cout << fixed << setprecision(10);
  }
} ___;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << "  ";
  __f(comma + 1, args...);
}
signed main() {
  long long int __, _ = 0;
  cin >> __;
  for (; _++ < __;) {
    long long int a, b;
    cin >> a >> b;
    long long int dd = (a & b);
    cout << (a ^ dd) + (b ^ dd) << '\n';
  }
}

### Prompt

Your challenge is to write a CPP solution to 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>
#pragma GCC optimize("Ofast")
using namespace std;
struct _ {
  ios_base::Init i;
  _() {
    cin.sync_with_stdio(0);
    cin.tie(0);
    cout << fixed << setprecision(10);
  }
} ___;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << "  ";
  __f(comma + 1, args...);
}
signed main() {
  long long int __, _ = 0;
  cin >> __;
  for (; _++ < __;) {
    long long int a, b;
    cin >> a >> b;
    long long int dd = (a & b);
    cout << (a ^ dd) + (b ^ dd) << '\n';
  }
}

```