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.
//package Codeforces;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;


public class B {

    public static int getCount(int num, int i) {
        int now = 1<<i;
        int dob = now<<1;

        int sb = Math.floorDiv(num, dob);
        sb *= now;

        int quo = Math.floorDiv(num, now);
        if(quo%2 == 1) {
            sb += num % now;
            ++sb;
        }
        return sb;
    }

    public static void main (String[] Z) throws IOException {

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder op = new StringBuilder();
        StringTokenizer stz;

        int T = Integer.parseInt(br.readLine());

        while(T-- > 0) {

            stz = new StringTokenizer(br.readLine());
            int l = Integer.parseInt(stz.nextToken());
            int r = Integer.parseInt(stz.nextToken());

            int max = 0;
            for (int i = 0; i < 21; i++) {

                int low = getCount(l-1, i);
                int high = getCount(r, i);

                int curr = high - low;
                max = Math.max(max, curr);

            }

            int ans = (r-l + 1) - max;
            op.append(ans + "\n");




        }

        System.out.println(op);


        // END OF CODE
    }
}