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.
// https://www.codechef.com/viewsolution/53050028

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public final class B {
    FastReader in = new FastReader();

    String n() throws Exception {
        return in.next();
    }

    String nln() throws Exception {
        return in.nextLine();
    }

    int ni() throws Exception {
        return Integer.parseInt(in.next());
    }

    long nl() throws Exception {
        return Long.parseLong(in.next());
    }

    double nd() throws Exception {
        return Double.parseDouble(in.next());
    }

    class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception {
            br = new BufferedReader(new FileReader(s));
        }

        String next() throws Exception {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new Exception(e.toString());
                }
            }
            return st.nextToken();
        }

        String nextLine() throws Exception {
            String str;
            try {
                str = br.readLine();
            } catch (IOException e) {
                throw new Exception(e.toString());
            }
            return str;
        }
    }

    public int numZerosInPosition(int l, int r, int m) {
        int r1 = r+1;
        int cycleLeftIdx = (int)Math.ceil(l*1.0/m);
        int cycleRightIdx = (int)Math.floor(r1*1.0/m);
//            int zeros1 = Math.max(m * cycleLeftIdx - l - m / 2, 0);
//            int zeros2 = Math.min(r1 - m * cycleRightIdx, m / 2);
//            int zeros3 = (cycleRightIdx - cycleLeftIdx) * (m / 2);
//            return zeros1+zeros2+zeros3;
        if (cycleRightIdx >= cycleLeftIdx) {
            int zeros1 = Math.max(m * cycleLeftIdx - l - m / 2, 0);
            int zeros2 = Math.min(r1 - m * cycleRightIdx, m / 2);
            int zeros3 = (cycleRightIdx - cycleLeftIdx) * (m / 2);
            return zeros1+zeros2+zeros3;
        } else {
            int zeros2 = Math.min(r1 - m * cycleRightIdx, m / 2);
            int notzeros = Math.min(l - m * cycleRightIdx, m/2);
            return zeros2 - notzeros;
        }
    }

    public void run() throws Exception {
        int numTests = ni();
        for (int testIdx = 0; testIdx < numTests; testIdx++) {
            int l = ni();
            int r = ni();
            int m = 2;
            int minNumZeros = r+1-l;
            while (r >= m/2) {
                int curNumZeros = numZerosInPosition(l, r, m);
                if (curNumZeros < minNumZeros) {
                    minNumZeros = curNumZeros;
                }
                m *= 2;
            }
            System.out.println(minNumZeros);
        }
    }

    public static void main(String[] args) throws Exception {
        B ct = new B();
        ct.run();
    }
}