Problem: On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters. Each of the pupils will go with the speed equal to v1. To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than k people at the same time) and the speed equal to v2. In order to avoid seasick, each of the pupils want to get into the bus no more than once.

Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected. 

Input

The first line of the input contains five positive integers n, l, v1, v2 and k (1 ≤ n ≤ 10 000, 1 ≤ l ≤ 109, 1 ≤ v1 < v2 ≤ 109, 1 ≤ k ≤ n) — the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus. 

Output

Print the real number — the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.

Examples

Input

5 10 1 2 5


Output

5.0000000000


Input

3 6 1 2 1


Output

4.7142857143

Note

In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5.

Solution (in JAVA):

Solution:
import static java.lang.Math.*;
import static java.util.Arrays.*;
import java.util.*;

import java.io.*;

public class Main {
	
	int v1, v2, K;
	void solve() {
		int N = sc.nextInt();
		int l = sc.nextInt();
		v1 = sc.nextInt();
		v2 = sc.nextInt();
		K = sc.nextInt();
		
		
		double hi = l / v1 + 1;
		double lo = 0;
		for (int step = 0; step < 250; step++) {
			double mid = (hi + lo) / 2;
			if (possible(N, l, mid)) hi = mid; else lo = mid;
		}
		out.printf("%.12f%n", hi);
	}
	
	boolean possible(int n, double l, double target_time) {
		if (l / v1 <= target_time) return true;
		double elapsed = 0;
		
		while (l > 0) {
			if (n <= K) {
				elapsed += l / v2;
				l = 0;
				break;
			} 
			double tr = target_time - elapsed;
			double t1 = (l - v1 * tr) / (v2 - v1);
			double t = (l - v1 * (tr - t1)) * 2 / (v1 + v2);
			n -= K;
			l -= v1 * t;
			elapsed += t;
		}
		return elapsed <= target_time;
	}
	
	static void tr(Object... os) { System.err.println(deepToString(os)); }
	static void tr(int[][] as) { for (int[] a : as) tr(a); }

	void print(int[] a) {
		if (a.length > 0) out.print(a[0]);
		for (int i = 1; i < a.length; i++) out.print(" " + a[i]);
		out.println();
	}

	public static void main(String[] args) throws Exception {
		new Main().run();
	}

	MyScanner sc = null;
	PrintWriter out = null;
	public void run() throws Exception {
		sc = new MyScanner(System.in);
		out = new PrintWriter(System.out);
		for (;sc.hasNext();) {
			solve();
			out.flush();
		}
		out.close();
	}

	class MyScanner {
		String line;
		BufferedReader reader;
		StringTokenizer tokenizer;

		public MyScanner(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream));
			tokenizer = null;
		}
		public void eat() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					line = reader.readLine();
					if (line == null) {
						tokenizer = null;
						return;
					}
					tokenizer = new StringTokenizer(line);
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
		}
		public String next() {
			eat();
			return tokenizer.nextToken();
		}
		public String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		public boolean hasNext() {
			eat();
			return (tokenizer != null && tokenizer.hasMoreElements());
		}
		public int nextInt() {
			return Integer.parseInt(next());
		}
		public long nextLong() {
			return Long.parseLong(next());
		}
		public double nextDouble() {
			return Double.parseDouble(next());
		}
		public int[] nextIntArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++) a[i] = nextInt();
			return a;
		}
	}
}