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: /*
   * Author:- Tanmay
   */
  import java.io.*;
  import java.util.*;
  public class Main {
  	static int log2(int N){
          return (int)(Math.log(N) / Math.log(2));
      }
  	public static void main(String[] args) {
  		FastScanner sc = new FastScanner();
  		PrintWriter out = new PrintWriter(System.out);
  		int dp[][] = new int[200001][20];
  		for(int i = 1; i <= 200000; i++){
  	        int te= i, idx = 0;
  	        while(te != 0){
  	            dp[i][idx++] += (te%2);
  	            te/=2;
  	        }
  	        for(int j = 0; j < 20; j++){
  	            dp[i][j] += dp[i-1][j];
  	        }
  		}
  //		int t=1;	
  		int t=sc.nextInt();
  		outer:	
  		while(t-- >0) {
  			int l = sc.nextInt(), r = sc.nextInt();
  			int a[] = new int[20];
  			int max = 0;
  			for(int i = 0; i < 20; i++){
  	            max = Math.max(max, dp[r][i] - dp[l-1][i]);
  	        }
  	 
  	        out.println(r - l + 1 - max);
  		}
  		out.flush();
  		out.close();
  	}
  	static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		int [] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		long [] longArray(int n) {
  			long[] a=new long[n];
  			for(int i=0 ; i<n ; i++) a[i]=nextLong();
  			return a;
  		}
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  	}
  } 