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 Codeforces
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        int N=200000,b[][]=new int[N+1][25],i,j;
        for(i=1;i<=N;i++)
        for(j=0;j<25;j++)
        {
            b[i][j]=b[i-1][j];
            if(((i>>j)&1)==1) b[i][j]++;
        }

        int t=Integer.parseInt(bu.readLine());
        while(t-->0)
        {
            String s[]=bu.readLine().split(" ");
            int l=Integer.parseInt(s[0]),r=Integer.parseInt(s[1]);
            int ans=r-l+1;
            for(i=0;i<25;i++)
            {
                int tot=r-l+1;
                tot-=b[r][i]-b[l-1][i];
                ans=Math.min(ans,tot);
            }
            sb.append(ans+"\n");
        }
        System.out.print(sb);
    }
}