Problem: An e-commerce startup pitches to the investors to get funding. They have been functional for n weeks now and also have a website!

For each week they know the number of unique visitors during this week vi and the revenue ci. To evaluate the potential of the startup at some range of weeks from l to r inclusive investors use the minimum among the maximum number of visitors multiplied by 100 and the minimum revenue during this period, that is: 

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The truth is that investors have no idea how to efficiently evaluate the startup, so they are going to pick some k random distinct weeks li and give them to managers of the startup. For each li they should pick some ri ≥ li and report maximum number of visitors and minimum revenue during this period.

Then, investors will calculate the potential of the startup for each of these ranges and take minimum value of p(li, ri) as the total evaluation grade of the startup. Assuming that managers of the startup always report the optimal values of ri for some particular li, i.e., the value such that the resulting grade of the startup is maximized, what is the expected resulting grade of the startup? 

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 1 000 000).

The second line contains n integers vi (1 ≤ vi ≤ 107) — the number of unique visitors during each week.

The third line contains n integers ci (1 ≤ ci ≤ 107) —the revenue for each week.

Output

Print a single real value — the expected grade of the startup. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6. 

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.

Examples

Input

3 2
3 2 1
300 200 300


Output

133.3333333

Note

Consider the first sample.

If the investors ask for li = 1 onwards, startup will choose ri = 1, such that max number of visitors is 3 and minimum revenue is 300. Thus, potential in this case is min(3·100, 300) = 300.

If the investors ask for li = 2 onwards, startup will choose ri = 3, such that max number of visitors is 2 and minimum revenue is 200. Thus, potential in this case is min(2·100, 200) = 200.

If the investors ask for li = 3 onwards, startup will choose ri = 3, such that max number of visitors is 1 and minimum revenue is 300. Thus, potential in this case is min(1·100, 300) = 100.

We have to choose a set of size 2 equi-probably and take minimum of each. The possible sets here are : {200, 300},{100, 300},{100, 200}, effectively the set of possible values as perceived by investors equi-probably: {200, 100, 100}. Thus, the expected value is (100 + 200 + 100) / 3 = 133.(3).

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.List;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskE solver = new TaskE();
		solver.solve(1, in, out);
		out.close();
	}

	static class TaskE {
		public void solve(int testNumber, InputReader in, PrintWriter out) {
			int n = in.nextInt();
			int k = in.nextInt();
			int[] v = new int[n];
			for (int i = 0; i < n; i++) {
				v[i] = -100 * in.nextInt();
			}
			int[] c = new int[n];
			for (int i = 0; i < n; i++) {
				c[i] = in.nextInt();
			}

			QueueMin<Integer> q1 = new QueueMin<>();
			QueueMin<Integer> q2 = new QueueMin<>();
			Integer[] a = new Integer[n];
			for (int i = 0, j = 0; i < n; i++) {
				if (i == j) {
					q1.addLast(v[i]);
					q2.addLast(c[i]);
					++j;
				}
				while (j < n) {
					int cur = Math.min(-q1.min(), q2.min());
					int nxt = Math.min(Math.max(-q1.min(), -v[j]), Math.min(q2.min(), c[j]));
					if (nxt > cur) {
						q1.addLast(v[j]);
						q2.addLast(c[j]);
						++j;
					} else {
						break;
					}
				}
				a[i] = Math.min(-q1.min(), q2.min());
				q1.removeFirst();
				q2.removeFirst();
			}

			for (int i = n - 1; i >= 0; i--) a[i] = Math.min(Math.max(-v[i], i == n - 1 ? 0 : a[i + 1]), c[i]);
			Arrays.sort(a);
			double sum = 0;
			double res = 0;
			for (int i = 0; i <= n - k; i++) {
				double cur = (1 - sum) * k / (n - i);
				sum += cur;
				res += cur * a[i];
			}

			out.printf("%.7f\n", res);
		}

		static class QueueMin<E extends Comparable<E>> {
			List<E[]> s1 = new ArrayList<>();
			List<E[]> s2 = new ArrayList<>();

			E min(E a, E b) {
				return a.compareTo(b) < 0 ? a : b;
			}

			public E min() {
				if (s1.isEmpty())
					return s2.get(s2.size() - 1)[1];
				if (s2.isEmpty())
					return s1.get(s1.size() - 1)[1];
				return min(s1.get(s1.size() - 1)[1], s2.get(s2.size() - 1)[1]);
			}

			public void addLast(E x) {
				E minima = x;
				if (!s1.isEmpty()) {
					minima = min(minima, s1.get(s1.size() - 1)[1]);
				}
				s1.add((E[]) new Comparable[]{x, minima});
			}

			public E removeFirst() {
				if (s2.isEmpty()) {
					E minima = null;
					while (!s1.isEmpty()) {
						E x = s1.remove(s1.size() - 1)[0];
						minima = minima == null ? x : min(minima, x);
						s2.add((E[]) new Comparable[]{x, minima});
					}
				}
				return s2.remove(s2.size() - 1)[0];
			}

		}

	}

	static class InputReader {
		final InputStream is;
		final byte[] buf = new byte[1024];
		int pos;
		int size;

		public InputReader(InputStream is) {
			this.is = is;
		}

		public int nextInt() {
			int c = read();
			while (isWhitespace(c))
				c = read();
			int sign = 1;
			if (c == '-') {
				sign = -1;
				c = read();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res = res * 10 + c - '0';
				c = read();
			} while (!isWhitespace(c));
			return res * sign;
		}

		int read() {
			if (size == -1)
				throw new InputMismatchException();
			if (pos >= size) {
				pos = 0;
				try {
					size = is.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (size <= 0)
					return -1;
			}
			return buf[pos++] & 255;
		}

		static boolean isWhitespace(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

	}
}