#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T second(T a) {
  return a * a;
}
template <class T>
inline string tostring(T a) {
  ostringstream os("");
  os << a;
  return os.str();
}
template <typename T>
inline long long tolong(T a) {
  long long res;
  istringstream os(a);
  os >> res;
  return res;
}
template <typename T>
inline T gcd(T a, T b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
template <typename T>
inline T bigmod(T a, T b, T m) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0)
    return second(bigmod(a, b / 2, m)) % m;
  else
    return (a % m * bigmod(a, b - 1, m)) % m;
}
const int inf = (int)1e9 + 5;
const long long linf = (long long)1e16 + 5;
const long long modd = (long long)1e9 + 7;
const int mod = 10000007;
void pr(int f) {
  if (f == 1)
    cout << "YES\n";
  else
    cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
    long long a, b;
    cin >> a >> b;
    long long x = max(a, b);
    long long y = min(a, b);
    cout << (y ^ x) << endl;
  }
}

### Prompt

Please formulate 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;
template <typename T>
inline T second(T a) {
  return a * a;
}
template <class T>
inline string tostring(T a) {
  ostringstream os("");
  os << a;
  return os.str();
}
template <typename T>
inline long long tolong(T a) {
  long long res;
  istringstream os(a);
  os >> res;
  return res;
}
template <typename T>
inline T gcd(T a, T b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
template <typename T>
inline T bigmod(T a, T b, T m) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0)
    return second(bigmod(a, b / 2, m)) % m;
  else
    return (a % m * bigmod(a, b - 1, m)) % m;
}
const int inf = (int)1e9 + 5;
const long long linf = (long long)1e16 + 5;
const long long modd = (long long)1e9 + 7;
const int mod = 10000007;
void pr(int f) {
  if (f == 1)
    cout << "YES\n";
  else
    cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
    long long a, b;
    cin >> a >> b;
    long long x = max(a, b);
    long long y = min(a, b);
    cout << (y ^ x) << endl;
  }
}

```