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.*;
  import java.io.*;
  import java.lang.*;
  public class problemB{
      public static void main(String[] args) throws Exception{
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          int t=Integer.parseInt(br.readLine());
          StringBuilder sb=new StringBuilder();
          while(t-->0){
              String[]s=br.readLine().split(" ");
              int l=Integer.parseInt(s[0]);
              int r=Integer.parseInt(s[1]);
              int p=(int)(Math.log(r)/Math.log(2));
              int[]arr=new int[p+1];
              for(int i=0;i<=p;i++){
                  int g=(r+1)/(1<<i);
                  arr[i]=(g/2)*(1<<i);
                  if(g%2!=0&&(r+1)%(1<<i)!=0){
                      arr[i]+=(r+1)%(1<<i);
                  }
              }
              int x=l-1;  
              int q=(int)(Math.log(x)/Math.log(2));
              for(int i=0;i<=q;i++){
                   int g=(x+1)/(1<<i);
                   arr[i]-=(g/2)*(1<<i);
                   if(g%2!=0&&(x+1)%(1<<i)!=0){
                      arr[i]-=(x+1)%(1<<i);
                   }
              }
              int min=Integer.MAX_VALUE,total=r-l+1;
              for(int i=0;i<=p;i++){
                  min=Math.min(min,total-arr[i]);
              }
              sb.append(min+"\n");
          }
          System.out.println(sb);
      }
  }