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 Main {
  	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 nextDouble() {
  			return Double.parseDouble(next());
  		}
  
  		String nextLine() {
  			String str = "";
  			try {
  				str = br.readLine();
  			} catch (IOException e) {
  				e.printStackTrace();
  			}
  			return str;
  		}
  	}
  
  	public static FastReader obj = new FastReader();
  	public static PrintWriter out = new PrintWriter(System.out);
  
  	public static void sort(long[] a) {
  		ArrayList<Long> arr = new ArrayList<>();
  		for (int i = 0; i < a.length; i++)
  			arr.add(a[i]);
  		Collections.sort(arr);
  		for (int i = 0; i < arr.size(); i++)
  			a[i] = arr.get(i);
  	}
  
  	public static void revsort(long[] a) {
  		ArrayList<Long> arr = new ArrayList<>();
  		for (int i = 0; i < a.length; i++)
  			arr.add(a[i]);
  		Collections.sort(arr, Collections.reverseOrder());
  		for (int i = 0; i < arr.size(); i++)
  			a[i] = arr.get(i);
  	}
  
  	//Cover the small test cases like for n=1 .
  	public static class pair {
  		long a;
  		long b;
  
  		pair(long x, long y) {
  			a = x;
  			b = y;
  		}
  	}
  
  	public static long l() {
  		return obj.nextLong();
  	}
  
  	public static int i() {
  		return obj.nextInt();
  	}
  
  	public static long[] i(int n) {
  		long[] arr = new long[n];
  		for (int i = 0; i < n; i++)
  			arr[i] = l();
  		return arr;
  	}
  
  	public static void p(long val) {
  		System.out.println(val);
  	}
  
  	public static void p(String s) {
  		System.out.println(s);
  	}
  
  	public static void p(long[] arr) {
  		for (int i = 0; i < arr.length; i++) {
  			out.print(arr[i] + " ");
  		}
  		out.println();
  	}
  
  	public static void sortpair(ArrayList<pair> arr) {
  		//ascending just change return 1 to return -1 and vice versa to get descending.
  		//compare based on value of pair.a
  		arr.sort(new Comparator<pair>() {
  			public int compare(pair o1, pair o2) {
  				long val = o1.a - o2.a;
  				if (val == 0)
  					return 0;
  				else if (val > 0)
  					return 1;
  				else
  					return -1;
  			}
  		});
  	}
  
  	public static void main(String[] args) {
          int[][] num=new int[200001][20];
          for(int i=1;i<=200000;i++)
          {
          	for(int j=0;j<20;j++)
          	{
          	   int tpow=(1<<j);
          	   num[i][j]=num[i-1][j];
          	   if((i&tpow)>0)num[i][j]+=1;	
          	}
          }
  		int len = i();
  		while (len-- != 0) {
  			int l=obj.nextInt();
  			int r=obj.nextInt();
  			int tut=r-l+1;
  			int max=0;
  			for(int i=0;i<20;i++) {
  				//System.out.println((num[r][i]-num[l-1][i])+" "+i);
  				max=Math.max(max, num[r][i]-num[l-1][i]);
  			}
  			out.println(tut-max);
              
  		}
  		out.flush();
  	}
  }