#include <bits/stdc++.h>
const unsigned long long inf = 1e18;
const int range = 1e6;
const long long inff = 1e-12;
const int dx[] = {0, 1, -1, 0};
const int dy[] = {-1, 0, 0, 1};
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
using namespace std;
bool cmp(const pair<long long, long long>& a,
         const pair<long long, long long>& b) {
  return (a.second > b.second);
}
void vecp(vector<unsigned long long> v) {
  for (int i = 0; i < int((v).size()); i++) {
    cout << v[i] << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long k, t, x, y;
  cin >> t;
  while (t--) {
    cin >> x >> y;
    if (x == y) {
      cout << 0 << endl;
    } else {
      cout << min((x ^ y), (y ^ x)) << endl;
    }
  }
}

### Prompt

Generate 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 unsigned long long inf = 1e18;
const int range = 1e6;
const long long inff = 1e-12;
const int dx[] = {0, 1, -1, 0};
const int dy[] = {-1, 0, 0, 1};
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
using namespace std;
bool cmp(const pair<long long, long long>& a,
         const pair<long long, long long>& b) {
  return (a.second > b.second);
}
void vecp(vector<unsigned long long> v) {
  for (int i = 0; i < int((v).size()); i++) {
    cout << v[i] << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long k, t, x, y;
  cin >> t;
  while (t--) {
    cin >> x >> y;
    if (x == y) {
      cout << 0 << endl;
    } else {
      cout << min((x ^ y), (y ^ x)) << endl;
    }
  }
}

```