#include <bits/stdc++.h>
const long long N = 100010, INF = 0x3f3f3f3f;
inline long long read(void) {
  long long res, f = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') f = -1;
  res = c - 48;
  while ((c = getchar()) >= '0' && c <= '9') res = res * 10 + c - 48;
  return res * f;
}
signed main(void) {
  long long T = read();
  while (T--) {
    long long a = read(), b = read();
    long long ans = 0;
    for (long long i = 35; i >= 0; --i) {
      if ((a & (1ll << i)) == (b & (1ll << i))) {
        ;
      } else {
        ans |= (1ll << i);
      }
    }
    printf("%I64d\n", ans);
  }
  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 N = 100010, INF = 0x3f3f3f3f;
inline long long read(void) {
  long long res, f = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') f = -1;
  res = c - 48;
  while ((c = getchar()) >= '0' && c <= '9') res = res * 10 + c - 48;
  return res * f;
}
signed main(void) {
  long long T = read();
  while (T--) {
    long long a = read(), b = read();
    long long ans = 0;
    for (long long i = 35; i >= 0; --i) {
      if ((a & (1ll << i)) == (b & (1ll << i))) {
        ;
      } else {
        ans |= (1ll << i);
      }
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```