#include <bits/stdc++.h>
using namespace std;
template <class T1, class T2>
ostream& operator<<(ostream& out, const pair<T1, T2> p) {
  out << '(' << p.first << ',' << p.second << ')';
  return out;
}
template <class T1, class T2>
istream& operator>>(istream& in, pair<T1, T2>& p) {
  in >> p.first >> p.second;
  return in;
}
template <class T>
istream& operator>>(istream& in, vector<T>& v) {
  for (T& x : v) in >> x;
  return in;
}
template <class T>
ostream& operator<<(ostream& out, const vector<vector<T>>& v) {
  for (const vector<T>& x : v) out << x << '\n';
  return out;
}
template <class T>
ostream& operator<<(ostream& out, const vector<T>& v) {
  for (const T& x : v) out << x << ' ';
  return out;
}
long long gcd(long long a, long long b) {
  if (b > a) swap(a, b);
  return (b ? gcd(b, a % b) : a);
}
using ll = long long;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using tiii = pair<pair<int, int>, int>;
using vi = vector<int>;
using vl = vector<long long>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<long long>>;
const int h = 1000000007;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cerr.setstate(ios::failbit);
  int t;
  cin >> t;
  while (t--) {
    int a, b;
    cin >> a >> b;
    for (int i = (0); i < (32); i++) {
      if (((a >> i) & 1) and ((b >> i) & 1)) {
        a -= (1 << i);
        b -= (1 << i);
      }
    }
    cout << a + b << '\n';
  }
}

### 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 <class T1, class T2>
ostream& operator<<(ostream& out, const pair<T1, T2> p) {
  out << '(' << p.first << ',' << p.second << ')';
  return out;
}
template <class T1, class T2>
istream& operator>>(istream& in, pair<T1, T2>& p) {
  in >> p.first >> p.second;
  return in;
}
template <class T>
istream& operator>>(istream& in, vector<T>& v) {
  for (T& x : v) in >> x;
  return in;
}
template <class T>
ostream& operator<<(ostream& out, const vector<vector<T>>& v) {
  for (const vector<T>& x : v) out << x << '\n';
  return out;
}
template <class T>
ostream& operator<<(ostream& out, const vector<T>& v) {
  for (const T& x : v) out << x << ' ';
  return out;
}
long long gcd(long long a, long long b) {
  if (b > a) swap(a, b);
  return (b ? gcd(b, a % b) : a);
}
using ll = long long;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using tiii = pair<pair<int, int>, int>;
using vi = vector<int>;
using vl = vector<long long>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<long long>>;
const int h = 1000000007;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cerr.setstate(ios::failbit);
  int t;
  cin >> t;
  while (t--) {
    int a, b;
    cin >> a >> b;
    for (int i = (0); i < (32); i++) {
      if (((a >> i) & 1) and ((b >> i) & 1)) {
        a -= (1 << i);
        b -= (1 << i);
      }
    }
    cout << a + b << '\n';
  }
}

```