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.*;
public class Codeforces {


	// CHECK CONSTRAINTS ALWAYS EDGE CASE MISSED 
	
static int mod = 1000000007;	

public static void main(String[] args) {
     Scanner sc = new Scanner(System.in);
        while(sc.hasNext()) {
    	int t =  sc.nextInt();
   int a[][] = new int[200002][32];
   
   for(int i = 1 ; i <=200000 ; i++) {
	 int temp = i;
	   for(int j = 0; j<32 ; j++) {
		   if((temp & 1) == 1) {
			  a[i][j]= a[i-1][j]+1;  
		   }
		   else {
			   a[i][j] = a[i-1][j];
		   }
		
		   temp = temp>>1;
	   }
	   
   }
   
 
   
   


    	
    	while(t-->0) {
    int l =  sc.nextInt();
    int r =  sc.nextInt();
       int max = Integer.MIN_VALUE;
   for(int i = 0 ; i<32 ; i++) {
	 
	  max = Math.max(max, a[r][i] - a[l-1][i]);
	  }
   
   int temp = r-l+1;
   System.out.println(temp - max);
   
   
   
     
     
    
    		
    		
    		
    	
    		
    	}

     
     
}

}
 

 
 }




        




		


        




		