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 B {
  
  	private static final FastScanner fs = new FastScanner();
  	private static final int MOD = 1000000007;
  	private static final int MAX = 2*100000;
  	private static final int[][] mat = new int[32][MAX+1];
  	public static void main(String[] args) throws IOException {
  		int cases = fs.nextInt();
  		for (int i = 0; i < 32; i++) {
  			int del = 0;
  			for (int j = 1; j <= MAX; j++) {
  				if (getBit(j, i) == 0) del++;
  				mat[i][j] = del;
  			}
  		}
  		while (cases-- > 0)
  			solve();
  	}
  	static void solve() throws IOException {
  		int l = fs.nextInt();
  		int r = fs.nextInt();
  		int res = Integer.MAX_VALUE;
  		for (int i = 0; i < 32; i++) {
  			res = Math.min(res, mat[i][r] - mat[i][l-1]);
  		}
  
  		System.out.println(res);
  	}
  
  	static int getBit(int n, int position)
  	{
  		return (n >> position) & 1;
  	}
  	static int setBit(int n, int position)
  	{
  		return n | (1 << position);
  	}
  	static int unsetBit(int n, int position)
  	{
  		return n & ~(1 << position);
  	}
  
  	static class FastScanner {
  		//I don't know how this works lmao
  		public 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());
  		}
  
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  
  		int[] nextInts(int n) {
  			int[] a = new int[n];
  			for (int i = 0; i < n; i++) a[i] = nextInt();
  			return a;
  		}
  	}
  }