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.io.PrintWriter;
import java.util.Scanner;

public class Main {
    
    public static void main(String[] args) {
        
        var sc = new Scanner(System.in);
        var pw = new PrintWriter(System.out);
        
        int T = Integer.parseInt(sc.next());
        for(int t = 0; t < T; t++){
            int l = Integer.parseInt(sc.next());
            int r = Integer.parseInt(sc.next());
            
            int max = 0;
            if(l%2 == 1 && r%2 == 1){
                max = (r-l+2) / 2;
            }else if(l%2 == 1 || r%2 == 1){
                max = (r-l+1) / 2;
            }else if(l%2 == 0 && r%2 == 0){
                max = (r-l) / 2;
            }
            for(int i = 0; i < 19; i++){
                int a = 1 << i;
                int b = (2 << i) - 1;
                
                if(r < a || b < l) continue;
                
                if(l <= a && b <= r){
                    max = Math.max(b-a+1, max);
                }else if(a <= l && r <= b){
                    max = Math.max(r-l+1, max);
                }else if(r < b){
                    max = Math.max(r-a+1, max);
                }else if(l < a){
                    max = Math.max(b-l+1, max);
                }
            }
            pw.println(r-l+1-max);
        }
        pw.flush();
    }
}