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.
#include <bits/stdc++.h>
using namespace std;
int t;
int arr[300005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  for (int i = 1; i < 300000; i++) {
    arr[i + 1] = arr[i] ^ i;
  }
  cin >> t;
  for (int l = 1; l <= t; l++) {
    int a, b;
    cin >> a >> b;
    if (a == 1) {
      if (b == 1) {
        cout << 3 << "\n";
      } else {
        cout << 2 << "\n";
      }
    } else {
      int ans = a;
      if (arr[a] == b) {
        cout << a << "\n";
      } else if ((arr[a] ^ a) == b) {
        cout << a + 2 << "\n";
      } else {
        cout << a + 1 << "\n";
      }
    }
  }
  return 0;
}
