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.*;
  import java.util.*;
  public class qB {
  
  	public static void main(String[] args)throws IOException {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		int T = Integer.parseInt(br.readLine());
  		while(T-- > 0) {
  			StringTokenizer st = new StringTokenizer(br.readLine());
  			int L = Integer.parseInt(st.nextToken());
  			int R = Integer.parseInt(st.nextToken());
  			int Rbit = (int) (Math.log(R) / Math.log(2) + 1);
  //			System.out.println(Rbit);
  			int[] bitCount = new int[20];
  			for(int i = 0;i < Rbit;i++) {
  				int cycle = 1 << i;
  				int Ldist = (L % cycle);
  				int Rdist = (R % cycle) + 1;
  				int Rtemp = R;
  				int Ltemp = L;
  				if((L / cycle) % 2 == 1) {
  					bitCount[i] -= Ldist;
  				}
  				Ltemp -= Ldist;
  				
  				if((R / cycle) % 2 == 1) {
  					bitCount[i] += Rdist;
  				}
  				Rtemp -= Rdist;
  				
  				if(Rtemp < Ltemp) {
  					continue;
  				}
  				boolean flagL = (Ltemp / cycle) % 2 == 1;
  				boolean flagR = (Rtemp / cycle) % 2 == 1;
  				
  				if(flagL || flagR) {
  					bitCount[i] += ((Rtemp - Ltemp) / (2 * cycle) + 1) * cycle;
  				}
  				else {
  					bitCount[i] += (Rtemp - Ltemp) / (2 * cycle) * cycle;
  				}
  			}
  			
  //			System.out.println(Arrays.toString(bitCount));
  			int max = Integer.MIN_VALUE;
  			for(int x: bitCount) {
  				max = Math.max(x, max);
  			}
  			
  			System.out.println(R - L + 1 - max);
  		}
  
  	}
  
  }
  