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.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;
		}
	}
}