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: /* package codechef; // don't place package name! */
  
  import java.util.*;
  import java.lang.*;
  import java.io.*;
  
  public class globalRoundC {
      static int mod = (int) (1e9 + 7);
      static long count = 0;
  
      public static void main(String[] args) throws java.lang.Exception {
          Scanner s = new Scanner(System.in);
          int t = s.nextInt();
  
          int[][] dp = new int[200001][33];
          for (int i = 1; i <= 200000; i++) {
              for (int j = 0; j < 33; j++) {
                  int m = 1 << j;
                  if ((i & m) != 0)
                      dp[i][j] = dp[i - 1][j] + 1;
                  else
                      dp[i][j] = dp[i - 1][j];
                  // dp[i][j]
              }
          }
          // printmat(dp);
  
          while (t-- > 0) {
  
              int l = s.nextInt();
              int r = s.nextInt();
  
              int n = r - l + 1;
              // int[] arr = new int[2*100000 + 1];
              // int l = scn.nextInt();
              // int r = scn.nextInt();
              int max1 = 0;
  
              for (int j = 0; j <= 32; j++) {
                  // int currmax= 0;
                  // currmax = Math.abs(dp[l][j] - dp[r][j]);
                  // System.out.println(dp[l][j] + " " + d p [l-1][j] +  " "+ dp[r][j]);
                  max1 = Math.max(dp[r][j] - dp[l - 1][j], max1);
              }
              // max1= Math.max(currmax,max1);
              // }
              int del = n - max1;
              System.out.println(del);
  
              // long maxsetbit = 0;
              // for (int i = 0 ; i< arr.length; i++){
              // maxsetbit = Math.max(maxsetbit, arr[i]);
              // }
  
          }
      }
  
      public static void printmat(int[][] matrix) {
          for (int i = 0; i < matrix.length; i++) {
              for (int j = 0; j < matrix[0].length; j++) {
                  System.out.print(matrix[i][j] + " ");
              }
              System.out.println();
          }
      }
  
  }