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.io.PrintWriter;
  import java.util.StringTokenizer;
  
  public class b {
  	public static void main(String[] args) {
  		FastScanner scan=new FastScanner();
  		PrintWriter out=new PrintWriter(System.out);
  		
  		int[][] a=new int[31][200001];
  		for(int log=30;log>=0;log--) {
  			for(int i=0;i<=200000;i++) {
  				if((i&(1<<log))>0) a[log][i]++;
  			}
  		}
  		
  		for(int i=0;i<=30;i++) {
  			for(int j=1;j<=200000;j++) {
  				a[i][j]+=a[i][j-1];
  			}
  		}
  		
  		int t=scan.nextInt();
  		for(int tt=0;tt<t;tt++) {
  			int l=scan.nextInt(), r=scan.nextInt();
  			int res=Integer.MAX_VALUE;
  			for(int log=30;log>=0;log--) {
  				int ct=rangesum(l,r,a[log]);
  				res=Math.min(res,r-l+1-ct);
  			}
  			out.println(res);
  		}
  		out.close();
  	}
  	public static int rangesum(int l, int r, int[] a) {
  		return l==0?a[r]:a[r]-a[l-1];
  	}
  	static class FastScanner {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public FastScanner() {
  			try	{
  				br = new BufferedReader(new InputStreamReader(System.in));
  				st = new StringTokenizer(br.readLine());
  			} catch (Exception e){e.printStackTrace();}
  		}
  
  		public String next() {
  			if (st.hasMoreTokens())	return st.nextToken();
  			try {st = new StringTokenizer(br.readLine());}
  			catch (Exception e) {e.printStackTrace();}
  			return st.nextToken();
  		}
  
  		public int nextInt() {return Integer.parseInt(next());}
  
  		public long nextLong() {return Long.parseLong(next());}
  
  		public double nextDouble() {return Double.parseDouble(next());}
  
  		public String nextLine() {
  			String line = "";
  			if(st.hasMoreTokens()) line = st.nextToken();
  			else try {return br.readLine();}catch(IOException e){e.printStackTrace();}
  			while(st.hasMoreTokens()) line += " "+st.nextToken();
  			return line;
  		}
  	}
  }
  