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.io.*;
import java.lang.*;
public class problemB{
    public static void main(String[] args) throws Exception{
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int t=Integer.parseInt(br.readLine());
        StringBuilder sb=new StringBuilder();
        while(t-->0){
            String[]s=br.readLine().split(" ");
            int l=Integer.parseInt(s[0]);
            int r=Integer.parseInt(s[1]);
            int p=(int)(Math.log(r)/Math.log(2));
            int[]arr=new int[p+1];
            for(int i=0;i<=p;i++){
                int g=(r+1)/(1<<i);
                arr[i]=(g/2)*(1<<i);
                if(g%2!=0&&(r+1)%(1<<i)!=0){
                    arr[i]+=(r+1)%(1<<i);
                }
            }
            int x=l-1;  
            int q=(int)(Math.log(x)/Math.log(2));
            for(int i=0;i<=q;i++){
                 int g=(x+1)/(1<<i);
                 arr[i]-=(g/2)*(1<<i);
                 if(g%2!=0&&(x+1)%(1<<i)!=0){
                    arr[i]-=(x+1)%(1<<i);
                 }
            }
            int min=Integer.MAX_VALUE,total=r-l+1;
            for(int i=0;i<=p;i++){
                min=Math.min(min,total-arr[i]);
            }
            sb.append(min+"\n");
        }
        System.out.println(sb);
    }
}