#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long ret = 0, ff = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    ret = ret * 10 + (ch ^ 48);
    ch = getchar();
  }
  return ret * ff;
}
void write(long long x) {
  if (x < 0) {
    x = -x, putchar('-');
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + 48);
}
void writeln(long long x) { write(x), puts(""); }
void writesp(long long x) { write(x), putchar(' '); }
signed main() {
  long long T = read();
  while (T--) {
    long long a = read(), b = read(), ans = 0;
    for (long long i = 0; i <= 40; i++) {
      long long x = (a >> i) & 1, y = (b >> i) & 1;
      if (x && y) ans |= (1ll << i);
    }
    writeln((a ^ ans) + (b ^ ans));
  }
  return 0;
}

### Prompt

Please formulate 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>
using namespace std;
inline long long read() {
  long long ret = 0, ff = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    ret = ret * 10 + (ch ^ 48);
    ch = getchar();
  }
  return ret * ff;
}
void write(long long x) {
  if (x < 0) {
    x = -x, putchar('-');
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + 48);
}
void writeln(long long x) { write(x), puts(""); }
void writesp(long long x) { write(x), putchar(' '); }
signed main() {
  long long T = read();
  while (T--) {
    long long a = read(), b = read(), ans = 0;
    for (long long i = 0; i <= 40; i++) {
      long long x = (a >> i) & 1, y = (b >> i) & 1;
      if (x && y) ans |= (1ll << i);
    }
    writeln((a ^ ans) + (b ^ ans));
  }
  return 0;
}

```