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.InputStreamReader;
  public class HelloWorld {
      public static void main(String [] args) throws java.io.IOException{
          BufferedReader scan = new BufferedReader(new InputStreamReader(System.in));
          int t = Integer.parseInt(scan.readLine());
          int[][] computed = preComputed();
          while(t-- > 0) {
              String[] lr = scan.readLine().split(" ");
              int l = Integer.parseInt(lr[0]);
              int r = Integer.parseInt(lr[1]);
              int min = Integer.MAX_VALUE;
              for(int i=0; i<31; i++) {
              	int toSub = ((l & (1 << i)) == 0) ?  computed[i][l]-1 : computed[i][l];
                  min = Math.min(min, computed[i][r]-toSub);
              }
              System.out.println(min);
          }
      }
      private static int[][] preComputed() {
          int[][] arr = new int[31][200001];
          for(int i=0; i<31; i++) {
              for(int j=1; j<200001; j++) {
                  int bit = j & (1 << i);
                  arr[i][j] = (bit == 0) ? arr[i][j-1]+1 : arr[i][j-1]; 
              }
          }
          return arr;
      }
  }