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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;

public class AndItsNonZero {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine().split(" ")[0]);
        PrintWriter pr = new PrintWriter(System.out);
        while (t > 0) {
            t--;
            String[] ints = br.readLine().split(" ");
            int l = Integer.parseInt(ints[0]);
            int r = Integer.parseInt(ints[1]);
            int rmaxx = 0;
            int lmaxx = 0;
            int x = 2;
            while(x<=(r*2)){
                int count = (r/x)*(x/2);
                int val = r%x+1-x/2;
                count += Integer.max(0,val);
                int count2 = (l-1)/x*(x/2);
                count2+= Integer.max(0,(l-1)%x + 1 - x/2);
                rmaxx = Integer.max(rmaxx, count - count2);
                x*=2;
            }
            x = 2;
            System.out.println(r - l + 1 - rmaxx);
            //found max m that is contained.

        }
        pr.close();
    }
}
