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;
long long x[300005];
int main() {
  long long t;
  cin >> t;
  long long xr;
  for (long long i = 0; i < 300005; i++) {
    if (i == 0) {
      xr = 0;
    } else {
      xr = xr ^ i;
    }
    x[i] = xr;
  }
  while (t--) {
    long long a, b;
    cin >> a >> b;
    xr = x[a - 1];
    if (xr == b) {
      cout << a << endl;
    } else if (xr == 0) {
      if (a == b) {
        cout << a + 2 << endl;
      } else
        cout << a + 1 << endl;
    } else {
      if ((xr ^ b) == (xr - b)) {
        long long d = xr - b;
        if (d == a) {
          cout << a + 2 << endl;
        } else
          cout << a + 1 << endl;
      } else {
        if ((xr & b) == xr) {
          if ((xr + b) == a) {
            cout << a + 2 << endl;
          } else
            cout << a + 1 << endl;
        } else {
          if ((xr ^ b) == a) {
            cout << a + 2 << endl;
          } else
            cout << a + 1 << endl;
        }
      }
    }
  }
}
