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.util.*;
  import java.io.*;
  
  public class B {
      static MyScanner sc;
      static PrintWriter out;
      static int[][] pref;
      static {
          sc = new MyScanner();
          out = new PrintWriter(System.out);
          int n = (int)2e5;
          pref = new int[n + 1][30];
          for(int j = 0; j < 30; j++) {
              for(int i = 1; i <= n; i++) {
                  pref[i][j] = pref[i - 1][j] + ((i >> j) & 1);
              }
          }
      }
      public static void solve() {
          int l = sc.nextInt();
          int r = sc.nextInt();
          int ans = 0;
          for(int i = 0; i < 30; i++)
              ans = Math.max(ans, pref[r][i] - pref[l - 1][i]);
          out.println(r - l + 1 - ans);
      }
      public static void main(String[] args) {
          int t = sc.nextInt();
          while(t-- > 0)
              solve();
          out.flush();
      }
  }
  
  class MyScanner {
      BufferedReader br;
      StringTokenizer tok;
      MyScanner() {
          try { br = new BufferedReader(new InputStreamReader(System.in)); }
          catch(Exception e) { System.out.println(e); }
          tok = new StringTokenizer("");
      }
      public String next() {
          try {
              while(!tok.hasMoreTokens()) tok = new StringTokenizer(br.readLine());
          }
          catch(Exception e) { System.out.println(e); }
          return tok.nextToken();
      }
      public int nextInt() {
          return Integer.parseInt(next());
      }
      public long nextLong() {
          return Long.parseLong(next());
      }
      public double nextDouble() {
          return Double.parseDouble(next());
      }
  }
  