#include <bits/stdc++.h>
using namespace std;
const int mod = (int)1000000000 + 7;
const int MOD = (int)998244353;
const int INF = (int)1e18;
bool cmp(const pair<int, int>& a, const pair<int, int>& b) {
  return a.second - a.first < b.second - b.first;
}
long long cc[27] = {0};
bool p[4000005];
vector<long long> prime;
void sieve() {
  for (long long i = 0; i < 4000005; i++) {
    p[i] = 1;
  }
  p[0] = p[1] = 0;
  for (long long i = 2; i < 4000005; i++) {
    if (p[i] == 1)
      for (long long j = i * i; j < 4000005; j += i) {
        p[j] = 0;
      }
  }
  for (long long i = 0; i < 4000005; i++) {
    if (p[i] == 1) prime.push_back(i);
  }
}
void solve() {
  long long x, y;
  cin >> x >> y;
  long long p = 0, i = 1;
  while (i <= max(x, y)) {
    if (x & i) {
      if (y & i) p += i;
    }
    i = i * 2;
  }
  cout << (x ^ p) + (y ^ p) << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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 int mod = (int)1000000000 + 7;
const int MOD = (int)998244353;
const int INF = (int)1e18;
bool cmp(const pair<int, int>& a, const pair<int, int>& b) {
  return a.second - a.first < b.second - b.first;
}
long long cc[27] = {0};
bool p[4000005];
vector<long long> prime;
void sieve() {
  for (long long i = 0; i < 4000005; i++) {
    p[i] = 1;
  }
  p[0] = p[1] = 0;
  for (long long i = 2; i < 4000005; i++) {
    if (p[i] == 1)
      for (long long j = i * i; j < 4000005; j += i) {
        p[j] = 0;
      }
  }
  for (long long i = 0; i < 4000005; i++) {
    if (p[i] == 1) prime.push_back(i);
  }
}
void solve() {
  long long x, y;
  cin >> x >> y;
  long long p = 0, i = 1;
  while (i <= max(x, y)) {
    if (x & i) {
      if (y & i) p += i;
    }
    i = i * 2;
  }
  cout << (x ^ p) + (y ^ p) << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```