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.
import java.util.*;

public class B {
  static final int MAXN = (int)3e5 + 5;
  static int[] f = new int[MAXN];
  
  public static void main(String[] args) {
    int t;
    Scanner sc = new Scanner(System.in);
    t = sc.nextInt();
    
    f[0] = 0;
    for (int i = 1; i < MAXN; ++i) {
      f[i] = f[i-1] ^ i;
    }
    
    while(t > 0) {
      t--;
      int n = sc.nextInt();
      int m = sc.nextInt();

      if (n > 0 && f[n-1] == m) {
        System.out.println(n);
      } else if ((f[n-1] ^ m) == n) {
        System.out.println(n+2);
      } else {
        System.out.println(n+1);
      }

    }
  }
}