At first, let's define function f(x) as follows: $$$ \begin{matrix} f(x) & = & \left\{ \begin{matrix} x/2 & \mbox{if } x  is even \\\ x - 1 & \mbox{otherwise } \end{matrix} \right. \end{matrix} $$$

We can see that if we choose some value v and will apply function f to it, then apply f to f(v), and so on, we'll eventually get 1. Let's write down all values we get in this process in a list and denote this list as path(v). For example, path(1) = [1], path(15) = [15, 14, 7, 6, 3, 2, 1], path(32) = [32, 16, 8, 4, 2, 1].

Let's write all lists path(x) for every x from 1 to n. The question is next: what is the maximum value y such that y is contained in at least k different lists path(x)?

Formally speaking, you need to find maximum y such that \left| \{ x ~|~ 1 ≤ x ≤ n, y ∈ path(x) \} \right| ≥ k.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 10^{18}).

Output

Print the only integer — the maximum value that is contained in at least k paths.

Examples

Input


11 3


Output


5


Input


11 6


Output


4


Input


20 20


Output


1


Input


14 5


Output


6


Input


1000000 100


Output


31248

Note

In the first example, the answer is 5, since 5 occurs in path(5), path(10) and path(11).

In the second example, the answer is 4, since 4 occurs in path(4), path(5), path(8), path(9), path(10) and path(11).

In the third example n = k, so the answer is 1, since 1 is the only number occuring in all paths for integers from 1 to 20.
import java.io.*;
import java.util.*;

public class C {

	static boolean occ(int x, int curr) {
		while (curr > 0) {
			if (curr == x)
				return true;
			if (curr % 2 == 1)
				curr--;
			else
				curr >>= 1;
		}
		return false;

	}

	static long cnt(long x, long n, boolean even) {
		if (x > n)
			return 0;
		long ans = 0;
		for (long m = 1, times = even ? 2 : 1; m <= n / x; m <<= 1, times <<= 1) {
			long currCnt = Math.min(times, n - m * x + 1);
			ans += currCnt;
		}
		return ans;

	}

	static long bs(long lo, long hi, int add, long k, long n) {
		long ans = 0;
		while (lo <= hi) {
			long mid = lo + hi >> 1;
			long cnt = cnt(2 * mid + add, n, add == 0);
			if (cnt >= k) {
				ans = mid;
				lo = mid + 1;
			} else
				hi = mid - 1;
		}
		return ans * 2 + add;
	}

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner();
		PrintWriter out = new PrintWriter(System.out);
		long n = sc.nextLong(), k = sc.nextLong();
		long lo = 1, hi = n / 2;
		long ans = bs(lo, hi, 0, k, n);
//		long ans=0;
		lo = 0;
		ans = Math.max(ans, bs(lo, hi, 1, k, n));
		out.println(ans);
		out.close();

	}

	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

		Scanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		Scanner(String fileName) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(fileName));
		}

		String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		String nextLine() throws IOException {
			return br.readLine();
		}

		int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		long nextLong() throws NumberFormatException, IOException {
			return Long.parseLong(next());
		}

		double nextDouble() throws NumberFormatException, IOException {
			return Double.parseDouble(next());
		}

		boolean ready() throws IOException {
			return br.ready();
		}

	}

	static void sort(int[] a) {
		shuffle(a);
		Arrays.sort(a);
	}

	static void shuffle(int[] a) {
		int n = a.length;
		Random rand = new Random();
		for (int i = 0; i < n; i++) {
			int tmpIdx = rand.nextInt(n);
			int tmp = a[i];
			a[i] = a[tmpIdx];
			a[tmpIdx] = tmp;
		}
	}

}