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;
  import java.io.PrintWriter;
  import java.util.StringTokenizer;
  
  public class B {
  
      void solve() throws IOException {
          int t = nextInt();
          /*
              EEENE
           */
          int[][] prefixSum = new int[19][200002];
          for (int i = 0; i < 19; ++i) {
              for (int j = 1; j <= 200000; ++j) {
                  int bitStatus = (j >> i) & 1;
                  bitStatus = bitStatus == 0 ? 1 : 0;
                  prefixSum[i][j + 1] = prefixSum[i][j] + bitStatus;
              }
          }
          for (int tt = 0; tt < t; ++tt) {
              int l = nextInt();
              int r = nextInt();
              int minDel = Integer.MAX_VALUE;
              for (int i = 0; i < 19; ++i) {
                  int zerosCount = prefixSum[i][r + 1] - prefixSum[i][l];
                  if (zerosCount > 0) {
                      minDel = Math.min(minDel, zerosCount);
                  } else {
                      minDel = 0;
                  }
              }
              writer.println(minDel);
          }
      }
  
      String nextToken() throws IOException {
          while (tokenizer == null || !tokenizer.hasMoreTokens()) {
              tokenizer = new StringTokenizer(reader.readLine());
          }
          return tokenizer.nextToken();
      }
  
      int nextInt() throws IOException {
          return Integer.parseInt(nextToken());
      }
  
      StringTokenizer tokenizer;
      PrintWriter writer;
      BufferedReader reader;
  
      public void run() {
          try {
              writer = new PrintWriter(System.out);
              reader = new BufferedReader(new InputStreamReader(System.in));
              solve();
              reader.close();
              writer.close();
          } catch (Exception e) {
              e.printStackTrace();
          }
      }
  
      public static void main(String[] args) {
          new B().run();
      }
  }