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 B {
    static ArrayList<Long> ls;

    public static void main(String args[]) {
        Scanner in = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);

        int t = in.nextInt();
        outer:
        for (int tt = 0; tt < t; tt++) {
            long l = in.nextLong();
            long r = in.nextLong();

            long max = 0;
            for (int i = 0; i < 32; i++) {
                long low = pow(2, i);
                long high = pow(2, i + 1);
                if (low < l) continue;
                if (low <= r && high <= r) {
                    long val = high - low;
                    max = Math.max(val, max);
                    continue;
                }
                if (low <= r) {
                    long val = r - low + 1;
                    max = Math.max(val, max);
                }
            }
            long ans = ((r - l) + 1) - max;
          //  debug(max);
            long temo = l;
            if (l > 1) temo--;
            long evL = temo / 2;
            long evH = r / 2;
            long odd = (evH - evL);
            // debug(odd, evH);
            ans = Math.min(ans, odd);
            pw.println(ans);
        }
        pw.close();
    }

    static long pow(long b, long p) {
        long res = 1;
        for (int i = 0; i < p; i++) res *= b;
        return res;
    }

    static void debug(Object... obj) {
        System.err.println(Arrays.deepToString(obj));
    }
}