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 {
    static int[][] bin;

    static void binar() {
        for (int i = 1; i < 200001; i++) {
            String s = Long.toBinaryString(i);
            for (int j = s.length() - 1; j >= 0; j--) {
                bin[i][s.length() - j - 1] = s.charAt(j) - '0' + bin[i - 1][s.length() - j - 1];
            }
        }
    }

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        //br = new BufferedReader(new FileReader("pencils.in"));
        //out = new PrintWriter("pencils.out");
        int t = sc.nextInt();
        bin = new int[200001][64];
        binar();
        for (int i = 0; i < t; i++) {
            int l = sc.nextInt();
            int r = sc.nextInt();
            int min = 2 * 100000 + 1;
            String d = Long.toBinaryString(l);
            for (int j = 0; j <= d.length(); j++) {
                min = Math.min(min, r - l + 1 - bin[r][j] + bin[l - 1][j]);
            }
            System.out.println(min);
        }
    }
}