#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const double PI = acos(-1.0);
int m, n, x, t, y, a[200005], k, l, r, z, all, ans, b[200005];
char c;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n, long long p = (long long)1000000007) {
  long long r = 1 % p;
  for (a %= p; n; a = a * a % p, n >>= 1)
    if (n & 1) r = r * a % p;
  return r;
}
long long inv(long long x, long long p = (long long)1000000007) {
  return x <= 1 ? 1 : inv(p % x, p) * (p - p / x) % p;
}
void init() {}
void solve() {
  scanf("%d%d", &x, &y);
  printf("%d\n", x ^ y);
}
int main() {
  int T;
  T = 1;
  scanf("%d", &T);
  for (int i = 1; i <= T; i++) {
    init();
    solve();
  }
  return 0;
}

### Prompt

Your task is to create 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;
const long long mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const double PI = acos(-1.0);
int m, n, x, t, y, a[200005], k, l, r, z, all, ans, b[200005];
char c;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n, long long p = (long long)1000000007) {
  long long r = 1 % p;
  for (a %= p; n; a = a * a % p, n >>= 1)
    if (n & 1) r = r * a % p;
  return r;
}
long long inv(long long x, long long p = (long long)1000000007) {
  return x <= 1 ? 1 : inv(p % x, p) * (p - p / x) % p;
}
void init() {}
void solve() {
  scanf("%d%d", &x, &y);
  printf("%d\n", x ^ y);
}
int main() {
  int T;
  T = 1;
  scanf("%d", &T);
  for (int i = 1; i <= T; i++) {
    init();
    solve();
  }
  return 0;
}

```