#include <bits/stdc++.h>
const long long int mod = 1e9 + 7;
using namespace std;
long long int pow(long long int x, long long int y) {
  long long int res = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
long long int modinv(long long int a) { return pow(a, mod - 2); }
void solution() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int a, b, x, y, i;
    cin >> a >> b;
    x = min(a, b);
    y = max(a, b);
    long long int mn = LONG_MAX;
    for (i = x; i <= y; i++) {
      long long int temp = ((a ^ i) + (b ^ i));
      if (temp > mn) break;
      mn = min(mn, temp);
    }
    cout << mn << "\n";
  }
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solution();
  return 0;
}

### 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>
const long long int mod = 1e9 + 7;
using namespace std;
long long int pow(long long int x, long long int y) {
  long long int res = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
long long int modinv(long long int a) { return pow(a, mod - 2); }
void solution() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int a, b, x, y, i;
    cin >> a >> b;
    x = min(a, b);
    y = max(a, b);
    long long int mn = LONG_MAX;
    for (i = x; i <= y; i++) {
      long long int temp = ((a ^ i) + (b ^ i));
      if (temp > mn) break;
      mn = min(mn, temp);
    }
    cout << mn << "\n";
  }
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solution();
  return 0;
}

```