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 int mod = (int) 1e9 + 7;

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

        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        int T = Integer.parseInt(bufferedReader.readLine());
        while (T-- > 0) {

            StringTokenizer stringTokenizer = new StringTokenizer(bufferedReader.readLine());
            int a = Integer.parseInt(stringTokenizer.nextToken());
            int b = Integer.parseInt(stringTokenizer.nextToken());

            long temp = count2(b);
            long temp1 = b - power(2, temp) + 1;
            long temp2 = power(2, temp) - power(2, temp - 1);

            long temp3 = (b - a + 1) % 2 != 0 ? (b - a + 1) / 2 + 1 : (b - a + 1) / 2;
            long ans = Math.max(temp1, Math.max(temp2, temp3));
            out.println(b - a + 1 - ans);

        }
        out.close();
    }

    private static long count2(long n) {

        int count = -1;
        while (n != 0) {
            n /= 2;
            count++;
        }
        return count;
    }

    public static long power(long a, long p) {
        long res = 1;
        while (p > 0) {
            if ((p & 1) == 1)
                res = (res * a) % mod;
            a = (a * a) % mod;
            p >>= 1;
        }
        return res;
    }

    static long gcd(long a, long b) {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    static long lcm(int a, int b) {
        return (a / gcd(a, b)) * b;
    }
}

/*
 * boolean[] sieve = new boolean[1_000_001];
 * Arrays.fill(sieve, true);
 * sieve[0] = false;
 * sieve[1] = false;
 * 
 * for (int x = 2; x * x < sieve.length; x++) {
 * if (sieve[x]) {
 * for (int j = x; j * x < sieve.length; j++) {
 * sieve[j * x] = false;
 * }
 * }
 * }
 */
