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 Main {
      public static void main(String args[]) {
          Scanner s = new Scanner(System.in);
          int t = s.nextInt();
          int[][] dp = new int[200001][20];
          for (int i = 0; i < 200001; i++){
              int check = i, idx = 0;
              while (check != 0){
                  dp[i][idx++] += (check%2); 
                  check /= 2;
              }
              for (int j = 0; j < 20; j++){
                  dp[i][j] += i > 0 ? dp[i-1][j] : 0;
              }
          }
          while (t-- > 0){
              int l = s.nextInt();
              int r = s.nextInt();
              int[] check = new int[20];
              int max = 0;
              for (int i = 0; i < 20; i++){
                  max = Math.max(max, dp[r][i]-dp[l-1][i]);
              }
              System.out.println((r-l+1)-max);
          }
      }
  }