#include <bits/stdc++.h>
using namespace std;
const long long MAX = 1e9;
const long long MOD = 1e9 + 7;
const long long pi = 3.1415926535897933846;
long long t, sum, best, worst, len, k, l, m, r, g, n, q, w, x, y, z, i, j, p;
vector<pair<long long, long long>> vp;
set<long long> sl;
unordered_map<long long, bool> bool_map;
vector<long long> v;
long long c[1000005];
long long dp[1000005] = {0};
string s;
long long seg[1000005] = {0};
long long prime[1000005] = {0};
void solve() {
  int x, y;
  cin >> x >> y;
  cout << (x ^ y) << "\n";
}
int main() {
  { ios_base::sync_with_stdio(false); }
  {
    cin.tie(NULL);
    cout.tie(NULL);
  }
  cin >> t;
  while (t--) solve();
}

### 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>
using namespace std;
const long long MAX = 1e9;
const long long MOD = 1e9 + 7;
const long long pi = 3.1415926535897933846;
long long t, sum, best, worst, len, k, l, m, r, g, n, q, w, x, y, z, i, j, p;
vector<pair<long long, long long>> vp;
set<long long> sl;
unordered_map<long long, bool> bool_map;
vector<long long> v;
long long c[1000005];
long long dp[1000005] = {0};
string s;
long long seg[1000005] = {0};
long long prime[1000005] = {0};
void solve() {
  int x, y;
  cin >> x >> y;
  cout << (x ^ y) << "\n";
}
int main() {
  { ios_base::sync_with_stdio(false); }
  {
    cin.tie(NULL);
    cout.tie(NULL);
  }
  cin >> t;
  while (t--) solve();
}

```