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.PrintWriter;
  import java.util.Scanner;
  
  public class Main {
      
      public static void main(String[] args) {
          
          var sc = new Scanner(System.in);
          var pw = new PrintWriter(System.out);
          
          var bit = new BinaryIndexedTree[18];
          for(int i = 0; i < 18; i++){
              bit[i] = new BinaryIndexedTree(200000);
          }
          for(int i = 1; i <= 200000; i++){
              String s = Integer.toBinaryString(i);
              int len = s.length();
              for(int j = 0; j < len; j++){
                  if(s.charAt(j) == '1'){
                      bit[len - 1 - j].add(i, 1);
                  }
              }
          }
          
          int T = Integer.parseInt(sc.next());
          for(int t = 0; t < T; t++){
              int l = Integer.parseInt(sc.next());
              int r = Integer.parseInt(sc.next());
              
              long max = 0;
              for(int i = 0; i < 18; i++){
                  max = Math.max(bit[i].sum(l, r+1), max);
              }
              pw.println(r-l+1-max);
          }
          pw.flush();
      }
      
      static class BinaryIndexedTree {
          int n;
          long[] bit;
          
          public BinaryIndexedTree(int n){
              this.n = n;
              bit = new long[n+1];
          }
          void add(int i, int x){
              while(i <= n){
                  bit[i] += x;
                  i += i & -i;
              }
          }
          long sum(int l, int r){
              return sum(r-1) - sum(l-1);
          }
          long sum(int i){
              long s = 0;
              while(i > 0){
                  s += bit[i];
                  i -= i & -i;
              }
              return s;
          }
      }
  }