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.Arrays;
  import java.util.StringTokenizer;
  
  public class Solution {
      static class Reader{
          private final BufferedReader reader;
          private StringTokenizer tokenizer;
          Reader(InputStream input)
          {
              this.reader = new BufferedReader(new InputStreamReader(input));
              this.tokenizer = new StringTokenizer("");
          }
          public String next() throws IOException {
              while(!tokenizer.hasMoreTokens())
              {
                  tokenizer = new StringTokenizer(reader.readLine());
              }
              return tokenizer.nextToken();
          }
          public int nextInt() throws IOException
          {
              return Integer.parseInt(next());
          }
          public double nextDouble() throws IOException
          {
              return Double.parseDouble(next());
          }
          public long nextLong() throws IOException
          {
              return Long.parseLong(next());
          }
  
          public int[] nextIntArr(int n) throws IOException {
              int[] arr = new int[n];
              for(int i=0;i<n;i++)arr[i]=nextInt();
              return arr;
          }
          public long[] nextLongArr(int n)throws IOException {
              long[] arr = new long[n];
              for(int i=0;i<n;i++)arr[i]=nextLong();
              return arr;
          }
          public char[] nextCharArr(int n) throws IOException {
              char[] arr = new char[n];
              for(int i=0;i<n;i++)arr[i] = next().charAt(0);
              return arr;
          }
      }
      public static void main(String[] args) throws IOException {
          Reader scan = new Reader(System.in);
          StringBuffer sb = new StringBuffer();
          BufferedOutputStream b = new BufferedOutputStream(System.out);
          int t = scan.nextInt();
          int[][] arr = new int[32][200001];
          for(int i=0;i<32;i++)
          {
              int pow2 = 1<<i;
              for(int j=1;j<200001;j++)
              {
                  if((j&pow2) == pow2)arr[i][j] = arr[i][j-1] + 1;
                  else arr[i][j] = arr[i][j-1];
              }
          }
          while(t-->0)
          {
              int l = scan.nextInt();
              int r = scan.nextInt();
              int min = Integer.MAX_VALUE;
              for(int i=0;i<32;i++)
              {
                  int x = arr[i][r] - arr[i][l-1] ;
                  min = Math.min(r-l+1 - x, min);
              }
              sb.append(min+"\n");
          }
          b.write((sb.toString()).getBytes());
          b.flush();
      }
  
  
  
  
  }
  