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 Main {
  	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  	static StringTokenizer st;
  	static PrintWriter pr = new PrintWriter(System.out);
  	static String readLine() throws IOException {
  		return br.readLine();
  	}
  	static String next() throws IOException {
  		while (st == null || !st.hasMoreTokens())
  			st = new StringTokenizer(readLine());
  		return st.nextToken();
  	}
  	static int readInt() throws IOException {
  		return Integer.parseInt(next());
  	}
  	static long readLong() throws IOException {
  		return Long.parseLong(next());
  	}
  	static double readDouble() throws IOException {
  		return Double.parseDouble(next());
  	}
  	static char readChar() throws IOException {
  		return next().charAt(0);
  	}
  	static class Pair implements Comparable<Pair> {
  		int f, s;
  		Pair(int f, int s) {
  			this.f = f;
  			this.s = s;
  		}
  		public int compareTo(Pair other) {
  			if (this.f != other.f) return this.f - other.f;
  			return this.s - other.s;
  		}
  	}
  	final static int inf = (int)1e9;
  	static int cnt(int x, int b) {
  		int n = 0;
  		if (x != 0) n = Integer.numberOfTrailingZeros(Integer.highestOneBit(x));
  		if (b > n) return 0;
  		int ret = (x>>(b + 1))*(1<<b);
  		if ((x>>b&1) == 0) return ret;
  		int add = 0;
  		for (int i = 0, mul = 1; i < b; ++i) {
  			add += mul*(x>>i&1);
  			mul <<= 1;
  		}
  		return ret + add + 1;
  	}
  	static void solve() throws IOException {
  		int L = readInt(), R = readInt(), num = R - L + 1, ans = num;
  		for (int b = 0; b < 20; ++b) {
  			ans = Math.min(ans, num - (cnt(R, b) - cnt(L - 1, b)));
  		}
  		pr.println(ans);
  	}
  	public static void main(String[] args) throws IOException {
  		//solve();
  		for (int t = readInt(); t > 0; --t) solve();
  		pr.close();
  	}
  }
  