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.util.Iterator;
  import java.util.StringTokenizer;
  
  public class AndItsNonZero {
  	
  	static class FastReader {
  		
          BufferedReader br;
          StringTokenizer st;
   
          public FastReader()
          {
              br = new BufferedReader(
                  new InputStreamReader(System.in));
          }
   
          String next()
          {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  }
                  catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
   
          int nextInt() { return Integer.parseInt(next()); }
   
          long nextLong() { return Long.parseLong(next()); }
          
          double nextFloat() { return Float.parseFloat(next()); }
   
          double nextDouble() { return Double.parseDouble(next()); }
   
          String nextLine()
          {
              String str = "";
              try {
                  str = br.readLine();
              }
              catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
          
          int[] nextArray(int n)
          {
          	int[] a = new int[n];
          	for (int i=0; i<n; i++) {
          		a[i] = this.nextInt();
          	}
          	return a;
          }
          
          long[] nextArrayLong(int n)
          {
          	long[] a = new long[n];
          	for (int i=0; i<n; i++) {
          		a[i] = this.nextLong();
          	}
          	return a;
          }
          
      }
  	
  	public static void solve(int l, int r, int[][] a) {
  		
  		int max = 0;
  		for (int i=0; i<20; i++) {
  			max = Math.max(max, a[i][r]-a[i][l-1]);
  		}
  		System.out.println(r-l-max+1);
  		
  	}
  
  	public static void main(String[] args) {
  		
  		FastReader in = new FastReader();
  		int[][] a = new int[20][200001];
  		for (int i=0; i<20; i++) {
  			int count = 0;
  			int mask = 1 << i;
  			for (int j=1; j<=200000; j++) {
  				if ((j & mask) != 0) {
  					count++;
  				}
  				a[i][j] = count;
  			}
  		}
  		int t = in.nextInt();
  		while (t-- > 0) {
  			solve(in.nextInt(), in.nextInt(), a);
  		}
  
  	}
  
  }
  