description: 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.
solution: 
  
  
  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);
     
     
     
       
       
      
      		
      		
      		
      	
      		
      	}
  
       
       
  }
  
  }
   
  
   
   }
  
  
  
  
          
  
  
  
  
  		
  
  
          
  
  
  
  
  		