#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 100;
long long v[40];
void solve() {
  memset(v, 0, sizeof(v));
  long long a, b;
  cin >> a >> b;
  long long i1 = 0, i2 = 0;
  long long cnt = 0;
  long long a1 = a, b1 = b;
  while (a1) {
    if (a1 % 2) ++v[cnt];
    a1 /= 2;
    ++cnt;
  }
  cnt = 0;
  while (b1) {
    if (b1 % 2) ++v[cnt];
    b1 /= 2;
    ++cnt;
  }
  long long f = 0;
  for (long long i = 0; i < 32; ++i) {
    if (v[i] == 2) i1 += (1 << i), f = 1;
  }
  if (!f)
    cout << a + b << endl;
  else
    cout << (a + b - i1 * 2) << endl;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
}

### Prompt

Create a solution in cpp for 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;
const int N = 2e5 + 100;
long long v[40];
void solve() {
  memset(v, 0, sizeof(v));
  long long a, b;
  cin >> a >> b;
  long long i1 = 0, i2 = 0;
  long long cnt = 0;
  long long a1 = a, b1 = b;
  while (a1) {
    if (a1 % 2) ++v[cnt];
    a1 /= 2;
    ++cnt;
  }
  cnt = 0;
  while (b1) {
    if (b1 % 2) ++v[cnt];
    b1 /= 2;
    ++cnt;
  }
  long long f = 0;
  for (long long i = 0; i < 32; ++i) {
    if (v[i] == 2) i1 += (1 << i), f = 1;
  }
  if (!f)
    cout << a + b << endl;
  else
    cout << (a + b - i1 * 2) << endl;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
}

```