#include <bits/stdc++.h>
using namespace std;
double EPS = 1e-9;
int INF = 1000000005;
long long INFF = 1000000000000000005LL;
double PI = acos(-1);
int dirx[8] = {-1, 0, 0, 1, -1, -1, 1, 1};
int diry[8] = {0, 1, -1, 0, -1, 1, -1, 1};
string decToBinary(int n) {
  string ans = "";
  int binaryNum[32];
  int i = 0;
  while (n > 0) {
    ans[i] = char(!(n % 2));
    n = n / 2;
    i++;
  }
  return ans;
}
void solve(int t) {
  long long a, b;
  cin >> a >> b;
  cout << (a ^ (a & b)) + (b ^ (a & b)) << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  long long t;
  cin >> t;
  for (int i = 1; i <= t; i++) solve(i);
}

### Prompt

Create a solution in CPP for 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;
double EPS = 1e-9;
int INF = 1000000005;
long long INFF = 1000000000000000005LL;
double PI = acos(-1);
int dirx[8] = {-1, 0, 0, 1, -1, -1, 1, 1};
int diry[8] = {0, 1, -1, 0, -1, 1, -1, 1};
string decToBinary(int n) {
  string ans = "";
  int binaryNum[32];
  int i = 0;
  while (n > 0) {
    ans[i] = char(!(n % 2));
    n = n / 2;
    i++;
  }
  return ans;
}
void solve(int t) {
  long long a, b;
  cin >> a >> b;
  cout << (a ^ (a & b)) + (b ^ (a & b)) << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  long long t;
  cin >> t;
  for (int i = 1; i <= t; i++) solve(i);
}

```