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 C {
  	
  	public static void main(String[] args)throws IOException {
  		
  		FastScanner scan = new FastScanner();
  		PrintWriter output = new PrintWriter(System.out);
  		int size = 200005;
  		int arr[][] = new int[size][20];
  		for(int i = 1;i<size;i++) {
  			char bits[] = Integer.toBinaryString(i).toCharArray();
  			int len = bits.length;
  			for(char j = 0;j<len;j++) {
  				if(bits[j]=='1') arr[i][len-j-1]++;
  			}
  		}
  		for(int i = 1;i<size;i++) {
  			for(int j = 0;j<20;j++) {
  				arr[i][j] += arr[i-1][j];
  			}
  		}
  		
  		int t = scan.nextInt();
  		for(int tt = 0;tt<t;tt++) {
  			int l = scan.nextInt(), r = scan.nextInt(), tot = r-l+1;
  			System.out.println(tot-findAns(arr, l, r));
  		}
  		output.flush();
  	}
  	
  	public static int findAns(int arr[][], int l, int r) {
  		int max = 0;
  		for(int i = 0;i<20;i++) {
  			if((arr[r][i]-arr[l-1][i])>max) max = arr[r][i]-arr[l-1][i];
  		}
  		return max;
  	}
  	
  	public static int[] sort(int arr[]) {
  		List<Integer> list = new ArrayList<>();
  		for(int i:arr) list.add(i);
  		Collections.sort(list);
  		for(int i = 0;i<list.size();i++) arr[i] = list.get(i);
  		return arr;
  	}
  	
  	public static int gcd(int a, int b) {
  		if(a == 0) return b;
  		return gcd(b%a, a);
  	}
  	
  	public static void printArray(int arr[]) {
  		for(int i:arr) System.out.print(i+" ");
  		System.out.println();
  	}
  	
  	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());
  		}
  		
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  	}
  
  }
  