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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  
  public class Main {
      public static void main(String[] args) throws IOException {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringBuilder sb = new StringBuilder();
          int[][] b = new int[200001][18];
          int n;
          for(int i=1;i<=200000;i++){
              n = i;
              for (int k = 17; k >= 0; k--) {
                  if (n >= (1 << k)) {
                      n %= (1 << k);
                      b[i][k] = b[i-1][k];
                  }else
                      b[i][k] = b[i-1][k] + 1;
              }
          }
          for (int T = Integer.parseInt(br.readLine()); T > 0; T--) {
              String[] s = br.readLine().split(" ");
              int i = Integer.parseInt(s[0]);
              int j = Integer.parseInt(s[1]);
              int min = Integer.MAX_VALUE;
              for(int k=0;k<18;k++)
                  min = Integer.min(min, b[j][k] - b[i-1][k]);
              sb.append(min).append('\n');
          }
          System.out.println(sb);
      }
  }