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.InputStreamReader;
public class HelloWorld {
    public static void main(String [] args) throws java.io.IOException{
        BufferedReader scan = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(scan.readLine());
        int[][] computed = preComputed();
        while(t-- > 0) {
            String[] lr = scan.readLine().split(" ");
            int l = Integer.parseInt(lr[0]);
            int r = Integer.parseInt(lr[1]);
            int min = Integer.MAX_VALUE;
            for(int i=0; i<31; i++) {
            	int toSub = ((l & (1 << i)) == 0) ?  computed[i][l]-1 : computed[i][l];
                min = Math.min(min, computed[i][r]-toSub);
            }
            System.out.println(min);
        }
    }
    private static int[][] preComputed() {
        int[][] arr = new int[31][200001];
        for(int i=0; i<31; i++) {
            for(int j=1; j<200001; j++) {
                int bit = j & (1 << i);
                arr[i][j] = (bit == 0) ? arr[i][j-1]+1 : arr[i][j-1]; 
            }
        }
        return arr;
    }
}