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 sfdfs
  {
  	public static void main(String[] args) {
  		Scanner sc=new Scanner(System.in);
  		int t=sc.nextInt();
  		
  		int arr[][]=new int[200001][32];
          for(int i=1;i<=200000;i++){
              for(int j=0;j<32;j++){
                  int mask=1<<j;
                  if((i&mask)!=0){
                      arr[i][j]=arr[i-1][j]+1;
                  }else{
                      arr[i][j]=arr[i-1][j];
                  }
              }
          }
  	        
  		for(int z=0;z<t;z++){
  		    int l=sc.nextInt();
  		    int r=sc.nextInt();
  	        
  	        int max=0;
  	        for(int i=0;i<32;i++){
  	            max=Math.max(max,arr[r][i]-arr[l-1][i]);
  	        }
  	        
  	        System.out.println((r-l+1)-max);
  		    
  		}
  	}
  }