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.OutputStream;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.PrintWriter;
  import java.util.StringTokenizer;
  import java.io.IOException;
  import java.io.BufferedReader;
  import java.io.InputStreamReader;
  import java.io.InputStream;
  
  /**
   * Built using CHelper plug-in
   * Actual solution is at the top
   */
  public class Main {
      public static void main(String[] args) {
          InputStream inputStream = System.in;
          OutputStream outputStream = System.out;
          FastScanner in = new FastScanner(inputStream);
          PrintWriter out = new PrintWriter(outputStream);
          TaskB solver = new TaskB();
          solver.solve(1, in, out);
          out.close();
      }
  
      static class TaskB {
          public void solve(int testNumber, FastScanner in, PrintWriter out) {
              int numTests = in.nextInt();
              for (int test = 0; test < numTests; test++) {
                  int l = in.nextInt();
                  int r = in.nextInt();
                  int ans = r - l;
                  for (int bit = 0; bit < 30; bit++) {
                      ans = Math.min(ans, f(r, bit) - f(l - 1, bit));
                  }
                  out.println(ans);
              }
          }
  
          private int f(int n, int i) {
              // Among the numbers [0, n], how many have bit zero in position i?
              int blockSize = 1 << i;
              int blockId = n / blockSize;
              int offset = n % blockSize;
              int res = (blockId + 1) / 2 * blockSize;
              if (blockId % 2 == 0) {
                  res += offset + 1;
              }
              return res;
          }
  
      }
  
      static class FastScanner {
          private BufferedReader in;
          private StringTokenizer st;
  
          public FastScanner(InputStream stream) {
              try {
                  in = new BufferedReader(new InputStreamReader(stream, "UTF-8"));
              } catch (Exception e) {
                  throw new AssertionError();
              }
          }
  
          public String next() {
              while (st == null || !st.hasMoreTokens()) {
                  try {
                      String rl = in.readLine();
                      if (rl == null) {
                          return null;
                      }
                      st = new StringTokenizer(rl);
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
              }
              return st.nextToken();
          }
  
          public int nextInt() {
              return Integer.parseInt(next());
          }
  
      }
  }
  
  