#include <bits/stdc++.h>
using namespace std;
int computeXOR(int n) {
  switch (n % 4) {
    case 0:
      return n;
      break;
    case 1:
      return 1;
      break;
    case 2:
      return n + 1;
      break;
    default:
      return 0;
      break;
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int a, b;
    int query = 0;
    cin >> a >> b;
    query += a;
    if (a - 1 >= 0 && computeXOR(a - 1) == b) {
      cout << query << "\n";
    } else if (a != (computeXOR(a - 1) ^ b)) {
      cout << query + 1 << "\n";
    } else {
      cout << query + 2 << "\n";
    }
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

Alice gave Bob two integers a and b (a > 0 and b ≥ 0). Being a curious boy, Bob wrote down an array of non-negative integers with \operatorname{MEX} value of all elements equal to a and \operatorname{XOR} value of all elements equal to b.

What is the shortest possible length of the array Bob wrote?

Recall that the \operatorname{MEX} (Minimum EXcluded) of an array is the minimum non-negative integer that does not belong to the array and the \operatorname{XOR} of an array is the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all the elements of the array.

Input

The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 5 ⋅ 10^4) — the number of test cases. The description of the test cases follows.

The only line of each test case contains two integers a and b (1 ≤ a ≤ 3 ⋅ 10^5; 0 ≤ b ≤ 3 ⋅ 10^5) — the \operatorname{MEX} and \operatorname{XOR} of the array, respectively.

Output

For each test case, output one (positive) integer — the length of the shortest array with \operatorname{MEX} a and \operatorname{XOR} b. We can show that such an array always exists.

Example

Input


5
1 1
2 1
2 0
1 10000
2 10000


Output


3
2
3
2
3

Note

In the first test case, one of the shortest arrays with \operatorname{MEX} 1 and \operatorname{XOR} 1 is [0, 2020, 2021].

In the second test case, one of the shortest arrays with \operatorname{MEX} 2 and \operatorname{XOR} 1 is [0, 1].

It can be shown that these arrays are the shortest arrays possible.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int computeXOR(int n) {
  switch (n % 4) {
    case 0:
      return n;
      break;
    case 1:
      return 1;
      break;
    case 2:
      return n + 1;
      break;
    default:
      return 0;
      break;
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int a, b;
    int query = 0;
    cin >> a >> b;
    query += a;
    if (a - 1 >= 0 && computeXOR(a - 1) == b) {
      cout << query << "\n";
    } else if (a != (computeXOR(a - 1) ^ b)) {
      cout << query + 1 << "\n";
    } else {
      cout << query + 2 << "\n";
    }
  }
}

```