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.*;
import java.lang.*;
import java.io.*;
public class Main {
    public static void main(String args[]) {
        Scanner s = new Scanner(System.in);
        int t = s.nextInt();
        while (t-- > 0){
            long l = s.nextLong();
            long r = s.nextLong();
            long n = pt(r);
            long ans1 = r-(long)Math.pow(2,n)+1;
            long ans2 = (long)Math.pow(2,n)-(long)Math.pow(2,n-1);
            long ans3 = ((r-l+1)%2) != 0 ? ((r-l+1)/2)+1 : (r-l+1)/2;
            long max = Math.max(ans1, Math.max(ans2, ans3));
            System.out.println((r-l+1)-max);
        }
    }

    public static long pt(long n){
        long cnt = -1;
        while (n!= 0){
            n /= 2;
            cnt++;
        }
        return cnt;
    }
}

/*
ll l, r;
    cin >> l >> r;
    ll pw = pt(r);
    ll ans1 = r - pow(2, pw) + 1;
    ll ans2 = pow(2, pw) - pow(2, pw - 1);
    ll mx = max(ans1 * 1LL, ans2 * 1LL);
    ll ans3 = ((r - l + 1) % 2) ? ((r - l + 1) / 2 + 1) : ((r - l + 1) / 2);
    mx = max(mx, ans3 * 1LL);
    cout << (r - l + 1) - mx << endl;
 
ll pt(ll n)
{
    ll cnt = -1;
    while (n != 0)
    {
        n /= 2;
        cnt++;
    }
    return cnt;
}
*/