#include <bits/stdc++.h>
using namespace std;
long long in() {
  char ch = getchar();
  int flag = 0;
  long long x = 0;
  while (ch < '0' || ch > '9') flag |= (ch == '-'), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + (ch - '0'), ch = getchar();
  return flag ? -x : x;
}
long long t, n, a, b, x[1000], y[1000];
int main() {
  t = in();
  while (t--) {
    a = in(), b = in();
    memset(x, 0, sizeof(x));
    memset(y, 0, sizeof(y));
    while (a) x[++x[0]] = a % 2, a /= 2;
    while (b) y[++y[0]] = b % 2, b /= 2;
    long long ans = 0;
    for (long long i = 1, k = 1; i <= max(x[0], y[0]); i++, k *= 2) {
      if (x[i] != y[i]) ans += k;
    }
    printf("%lld\n", ans);
  }
  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 namespace std;
long long in() {
  char ch = getchar();
  int flag = 0;
  long long x = 0;
  while (ch < '0' || ch > '9') flag |= (ch == '-'), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + (ch - '0'), ch = getchar();
  return flag ? -x : x;
}
long long t, n, a, b, x[1000], y[1000];
int main() {
  t = in();
  while (t--) {
    a = in(), b = in();
    memset(x, 0, sizeof(x));
    memset(y, 0, sizeof(y));
    while (a) x[++x[0]] = a % 2, a /= 2;
    while (b) y[++y[0]] = b % 2, b /= 2;
    long long ans = 0;
    for (long long i = 1, k = 1; i <= max(x[0], y[0]); i++, k *= 2) {
      if (x[i] != y[i]) ans += k;
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```