You are given an array consisting of all integers from [l, r] inclusive. For example, if l = 2 and r = 5, the array would be [2, 3, 4, 5]. What's the minimum number of elements you can delete to make the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the array non-zero?

A bitwise AND is a binary operation that takes two equal-length binary representations and performs the AND operation on each pair of the corresponding bits.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.

The first line of each test case contains two integers l and r (1 ≤ l ≤ r ≤ 2 ⋅ 10^5) — the description of the array.

Output

For each test case, output a single integer — the answer to the problem.

Example

Input


5
1 2
2 8
4 5
1 5
100000 200000


Output


1
3
0
2
31072

Note

In the first test case, the array is [1, 2]. Currently, the bitwise AND is 0, as 1\ \& \ 2 = 0. However, after deleting 1 (or 2), the array becomes [2] (or [1]), and the bitwise AND becomes 2 (or 1). This can be proven to be the optimal, so the answer is 1.

In the second test case, the array is [2, 3, 4, 5, 6, 7, 8]. Currently, the bitwise AND is 0. However, after deleting 4, 5, and 8, the array becomes [2, 3, 6, 7], and the bitwise AND becomes 2. This can be proven to be the optimal, so the answer is 3. Note that there may be other ways to delete 3 elements.

import java.util.*;
import java.lang.*;
import java.io.*;

public class HelloWorld{

    public static void main(String []args){
        Scanner s  = new Scanner(System.in);
        
        long factor = 1, templ, tempr, cnt;
        long i;
        int t = s.nextInt();
        
        while(t > 0) {
            cnt = 0;
            //System.out.println(t);
            long l = s.nextLong();
            long r = s.nextLong();
            
            templ = l;
            tempr = r;
            
            factor = 1;
            while(templ != tempr) {
                templ = templ / 2;
                tempr = tempr / 2;
                factor *= 2;
            }
            
            if (factor * templ != 0) {
                System.out.println(0);
                t--;
                continue;
            }
            
            long prev = l;
            for(i = l+1; i <= r; i = i + 1) {
                if((prev & i) == 0) {
                    cnt++;
                    continue;
                } else {
                    prev = (prev & i);
                }
            }
            
            System.out.println(cnt);
            
            t--;
        }
    }
}