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.
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());
	}
}