Problem: Ilya got tired of sports programming, left university and got a job in the subway. He was given the task to determine the escalator load factor. 

Let's assume that n people stand in the queue for the escalator. At each second one of the two following possibilities takes place: either the first person in the queue enters the escalator with probability p, or the first person in the queue doesn't move with probability (1 - p), paralyzed by his fear of escalators and making the whole queue wait behind him.

Formally speaking, the i-th person in the queue cannot enter the escalator until people with indices from 1 to i - 1 inclusive enter it. In one second only one person can enter the escalator. The escalator is infinite, so if a person enters it, he never leaves it, that is he will be standing on the escalator at any following second. Ilya needs to count the expected value of the number of people standing on the escalator after t seconds. 

Your task is to help him solve this complicated task.

Input

The first line of the input contains three numbers n, p, t (1 ≤ n, t ≤ 2000, 0 ≤ p ≤ 1). Numbers n and t are integers, number p is real, given with exactly two digits after the decimal point.

Output

Print a single real number — the expected number of people who will be standing on the escalator after t seconds. The absolute or relative error mustn't exceed 10 - 6.

Examples

Input

1 0.50 1


Output

0.5


Input

1 0.50 4


Output

0.9375


Input

4 0.20 2


Output

0.4

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.math.*;

public class D518 {

	public static BufferedReader read = null;
	public static PrintWriter out = null;
	public static StringTokenizer token = null;
	
	
	public static void solve()
	{
		int n = nint();
		double p = ndouble();
		int t = nint();
		// dp[i][j] = prob that j people are on after i seconds
		// 0 for j > i
		double dp[][] = new double[t+1][t+1];
		dp[1][0] = 1-p;
		dp[1][1] = p;
		for(int i=2; i<=t; i++)
		{
			dp[i][0] = dp[i-1][0] * (1-p);
			for(int j=1; j<=i; j++)
			{
				dp[i][j] = dp[i-1][j-1] * p + dp[i-1][j] * (1-p);
			}
		}
		//System.out.println(Arrays.toString(dp[t]));
		double e = 0;
		for(int i=0; i<=t; i++)
		{
			int mul = i;
			if(mul > n) mul = n; 
			e+=dp[t][i]*mul;
		}
		out.printf("%.15f", e);
		
	}
	
	public static void main(String[] args)
	{
		read = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		solve();
		out.flush();
		out.close();
		
	}
	
	// i/o functions
	public static String next() // returns the next string
	{
		while(token == null || !token.hasMoreTokens())
		{
			try {
				token = new StringTokenizer(read.readLine());
			} catch (IOException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
		}
		return token.nextToken();
	}
	
	public static int nint()
	{
		return Integer.parseInt(next());
	}
	public static long nlong()
	{
		return Long.parseLong(next());
	}
	public static double ndouble()
	{
		return Double.parseDouble(next());
	}
	public static int[] narr(int n)
	{
		int[] a = new int[n];
		for(int i=0; i<n; i++) a[i] = nint();
		return a;
	}
	
	public static long[] nal(int n)
	{
		long[] a = new long[n];
		for(int i=0; i<n; i++) a[i] = nlong();
		return a;
	}
	
	
}