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 AndItsNonZero{
    static int dp[][]=new int[20][200001];
    public static void main (String args[]) throws IOException{
        for(int i=0; i<20; i++)
            dp[i][0]=0;     // smallest possible value of l is 1
        for(int i=0; i<20; i++){
            for(int j=1; j<=200000; j++){
                dp[i][j]=dp[i][j-1]+(~(j>>i)&1);
            }
        }
        BufferedReader ob = new BufferedReader(new InputStreamReader(System.in));
        int t=Integer.parseInt(ob.readLine());
        while(t-->0){
            StringTokenizer st=new StringTokenizer(ob.readLine());
            int l=Integer.parseInt(st.nextToken());
            int r=Integer.parseInt(st.nextToken());
            int answer=Integer.MAX_VALUE;
            for(int i=0; i<20; i++){
                answer=Math.min(answer, dp[i][r]-dp[i][l-1]);
            }
            System.out.println(answer);
        }
    }
}