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.*;
import java.util.*;
public class Main{
    public static void main (String [] args) throws IOException{
        BufferedReader br = new BufferedReader (new InputStreamReader (System.in));
        int t = Integer.parseInt(br.readLine());
        while (t-->0){
            StringTokenizer st = new StringTokenizer(br.readLine());
            int l = Integer.parseInt(st.nextToken());
            int r = Integer.parseInt(st.nextToken());
            int ar[] = new int[33];
            Arrays.fill(ar,-1);
            int c=0,m=-1;
            for (int i=l;i<=r;i++){
                c=0;
                int x=i;
                while (x>0){
                    if ((x&1)==0) ar[c]+=1;
                    else{
                        if (ar[c]==-1) ar[c]=i-l;
                    }
                    x = (x>>1);
                    c++;
                }
                //if ((c-1)>m) m = c-1;
            }
            int ans = Integer.MAX_VALUE;
            for (int i=0;i<=32;i++){
                //System.out.print (ar[i]+" ");
                if (ar[i]<ans && ar[i]!=-1) ans = ar[i];
            }
            System.out.println (ans);
        }
    }
}