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.InputStreamReader;
  import java.io.PrintWriter;
  import java.math.BigDecimal;
  import java.math.BigInteger;
  import java.util.*;
  
  public class Main {
  	public static void main(String[] args) {
  		InputReader in = new InputReader();
  		PrintWriter out = new PrintWriter(System.out);
  		solve(in, out);
  
  		out.close();
  	}
  
  	public static void solve(InputReader in, PrintWriter out) {
  		int[][] pre = new int[20][200005];
  		for (int i = 0; i < 20; i++) {
  			for (int j = 1; j < 200005; j++) {
  				pre[i][j] = pre[i][j - 1] + ((j >> i) & 1);
  			}
  		}
  
  		// TODO
  		int t = in.nextInt();
  		while (t-- > 0) {
  			int l = in.nextInt(), r = in.nextInt();
  			int n = r - l + 1;
  			int res = n;
  			for (int i = 0; i < 20; i++) {
  				res = Math.min(res, n - (pre[i][r] - pre[i][l - 1]));
  			}
  			out.println(res);
  		}
  	}
  }
  
  // ctrl+shsift+p
  class InputReader {
  	BufferedReader buf;
  	StringTokenizer tok;
  
  	InputReader() {
  		buf = new BufferedReader(new InputStreamReader(System.in));
  	}
  
  	boolean hasNext() {
  		while (tok == null || !tok.hasMoreElements()) {
  			try {
  				tok = new StringTokenizer(buf.readLine());
  			} catch (Exception e) {
  				return false;
  			}
  		}
  		return true;
  	}
  
  	String next() {
  		if (hasNext()) return tok.nextToken();
  		return null;
  	}
  
  	int nextInt() {
  		return Integer.parseInt(next());
  	}
  
  	long nextLong() {
  		return Long.parseLong(next());
  	}
  
  	double nextDouble() {
  		return Double.parseDouble(next());
  	}
  
  	BigInteger nextBigInteger() {
  		return new BigInteger(next());
  	}
  
  	BigDecimal nextBigDecimal() {
  		return new BigDecimal(next());
  	}
  }