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.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();
  		
  	}
  
  }
  