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 {

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

		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		
		int[][] cnts = new int[200001][18];
		
		for(int i = 1; i <= 200000; i++) {
									
			for(int j = 0; j < 18; j++) {
				
				cnts[i][j] += cnts[i-1][j];
				
				if(((i>>j) & 1) == 1)
					cnts[i][j]++;
				
			}
			
		}
		
		int t = Integer.parseInt(br.readLine());
		
		while(t --> 0) {
			
			StringTokenizer st = new StringTokenizer(br.readLine());
			
			int l = Integer.parseInt(st.nextToken());
			int r = Integer.parseInt(st.nextToken());
			
			int mx = cnts[r][0] - cnts[l-1][0];
			
			for(int i = 1; i < 18; i++) {
				
				mx = Math.max(mx, cnts[r][i] - cnts[l-1][i]);
				
			}
			
			pw.println((1+r-l)-mx);
			
		}
		
		pw.close();
		
	}

}
