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

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb = new StringBuilder();
        int[][] b = new int[200001][18];
        int n;
        for(int i=1;i<=200000;i++){
            n = i;
            for (int k = 17; k >= 0; k--) {
                if (n >= (1 << k)) {
                    n %= (1 << k);
                    b[i][k] = b[i-1][k];
                }else
                    b[i][k] = b[i-1][k] + 1;
            }
        }
        for (int T = Integer.parseInt(br.readLine()); T > 0; T--) {
            String[] s = br.readLine().split(" ");
            int i = Integer.parseInt(s[0]);
            int j = Integer.parseInt(s[1]);
            int min = Integer.MAX_VALUE;
            for(int k=0;k<18;k++)
                min = Integer.min(min, b[j][k] - b[i][k]);
            sb.append(min).append('\n');
        }
        System.out.println(sb);
    }
}