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.Scanner;

/**
 * @author dzx
 * @data 2021/12/25 -13:21
 */
public class test {
    static int[][] a = new int[200001][31];
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int test_num = in.nextInt();
        for (int i = 0; i <test_num ; i++) {
            int c = in.nextInt();
            int b = in.nextInt();
            if(c==0){
                int min = Integer.MAX_VALUE;
                for (int j = 0; j < 31; j++) {
                    min = Math.min(a[b][j], min);
                }
                System.out.println(min);
            }else{
                int min = Integer.MAX_VALUE;
                for (int j = 0; j < 31; j++) {
                    min = Math.min(a[b][j]-a[c-1][j], min);
                }
                System.out.println(min);
            }
        }
    }
    static{
        for (int i = 1; i < 200001; i++) {
            for (int j = 0; j < 31; j++) {
                int num=1<<j;
                a[i][j] += (num & i) == 0 ? 1 : 0;
                a[i][j] += a[i - 1][j];
            }
        }
    }
}
