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.*;
  import java.util.*;
  public class B {
     public static void main(String[] hi) {
  	  int[][] dp = new int[200001][32];
  	  int[] ones = new int[32];
  	  for (int i=0; i<dp.length; i++) {
  
  		  String s = Integer.toBinaryString(i);
  		  int digit = 1;
  		  for (int k=s.length()-1; k>=0; k--) {
  			  if (s.charAt(k) == '1') {
  				  //System.out.println(ones[32-digit]);
  				  ones[32-digit] += 1;
  			  }
  			  digit++;
  		  }
  		  int[] copy = new int[ones.length];
  		  for (int k=0; k<ones.length; k++) {
  			  copy[k] = ones[k];
  		  }
  		  dp[i] = copy;
  		  //System.out.println(Arrays.toString(ones));
  	  }
  	  FastIO io = new FastIO();
        int t = io.nextInt();
        while (t --> 0) {
           int l = io.nextInt(), r = io.nextInt();         
           int[] array = new int[32];
           int max = 0;
           //System.out.println(Arrays.toString(dp[r]));
           //System.out.println(Arrays.toString(dp[l-1]));
           for (int i=0; i<array.length; i++) {
          	 array[i] = dp[r][i]-dp[l-1][i];
          	 
          	 max = Math.max(max, array[i]);
           }
           //System.out.println(Arrays.toString(array));
           io.println((r-l+1)-max);
        }
        io.close();
     }
     static class FastIO extends PrintWriter {
        BufferedReader br;
        StringTokenizer st;
        public FastIO() {
           super(System.out);
           br = new BufferedReader(new InputStreamReader(System.in));
        }
        public String next() {
           try {
              while (st == null || !st.hasMoreTokens()) {
                 st = new StringTokenizer(br.readLine());
              }
              return st.nextToken();
           } 
           catch (Exception e) {
              return null;
           }
        }
        public int nextInt() {
           return Integer.parseInt(next());
        }
        public double nextDouble() {
           return Double.parseDouble(next());
        }
        public long nextLong() {
           return Long.parseLong(next());
        }
        public void close() {
           try {
              br.close();
           } 
           catch (IOException e) {
              e.printStackTrace();
           }
           super.close();
        }
     }
     static class input extends PrintWriter {
        BufferedReader br;
        StringTokenizer st;
        public input() {
           super(System.out);
           br = new BufferedReader(new InputStreamReader(System.in));
        }
        public String next() {
           try {
              while (st == null || !st.hasMoreTokens()) {
                 st = new StringTokenizer(br.readLine());
              }
              return st.nextToken();
           } 
           catch (Exception e) {
              return null;
           }
        }
        public void close() {
           try {
              br.close();
           } 
           catch (IOException e) {
              e.printStackTrace();
           }
           super.close();
        }
     }
  }