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.util.Arrays;
import java.util.Scanner;

public class b
{
	public static void main(String[] args)
	{
		Scanner scanner = new Scanner (System.in);
		int numC = scanner.nextInt();
		
		int[][] pref = new int[200010][20];
		
		pref[1][0] = 1;
		
		for (int i = 2; i < pref.length; i++)
		{
			for (int j = 0; j < 20; j++)
			{
				pref[i][j] = pref[i - 1][j] + ((i & (1 << j)) == 0 ? 0 : 1);
			}
		}
		
		for (int c = 0; c < numC; c++)
		{
			int l = scanner.nextInt();
			int r = scanner.nextInt();
			
			int max = -1;
			for (int i = 0; i < 20; i++)
			{
				int tryThis = pref[r][i] - pref[l - 1][i];
				if (tryThis > max)
				{
					max = tryThis;
				}
			}
			
			System.out.println(r - l + 1 - max);
		
		}
	}
}
