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 {
      static long[][] calc;
      static int N = ((int) 1e5) * 3;
  
  
      public static void main(String args[]) {
          Scanner in = new Scanner(System.in);
          PrintWriter pw = new PrintWriter(System.out);
  
          build();
          int t = in.nextInt();
          for (int tt = 0; tt < t; tt++) {
              int l = in.nextInt(), r = in.nextInt();
              long total = r - l;
              total++;
              l--;
  
              long max = 0;
              for (int i = 0; i < 32; i++) {
                  max = Math.max(max, calc[r][i] - calc[l][i]);
              }
              pw.println((total - max));
          }
          pw.close();
      }
  
      static void build() {
          calc = new long[N][32];
          for (int i = 1; i < N; i++) {
              for (int j = 0; j < 32; j++) {
                  if ((i & (1 << j)) != 0) {
                      calc[i][j]++;
                  }
              }
          }
          for (int i = 1; i < N; i++) {
              for (int j = 0; j < 32; j++) calc[i][j] += calc[i - 1][j];
          }
      }
  
      static void debug(Object... obj) {
          System.err.println(Arrays.deepToString(obj));
      }
  }