GRAVITATION, n. “The tendency of all bodies to approach one another with a strength proportion to the quantity of matter they contain – the quantity of matter they contain being ascertained by the strength of their tendency to approach one another. This is a lovely and edifying illustration of how science, having made A the proof of B, makes B the proof of A.”

                                                                       Ambrose Bierce

You have a population of k Tribbles. This particular species of Tribbles live for exactly one day and then die. Just before death, a single Tribble has the probability Pi of giving birth to i more Tribbles. What is the probability that after m generations, every Tribble will be dead?

Input

The first line of input gives the number of cases, N. N test cases follow. Each one starts with a line containingn(1≤n≤1000),k(0≤k≤1000)andm(0≤m≤1000). Thenextnlineswillgivethe probabilities P0, P1, . . . , Pn−1.

Output

For each test case, output one line containing ‘Case #x:’ followed by the answer, correct up to an absolute or relative error of 10−6.

Sample Input

4 3 1 1 0.33 0.34 0.33 3 1 2 0.33 0.34 0.33 3 1 2 0.5 0.0 0.5 4 2 2 0.5 0.0 0.0 0.5

Sample Output

Case #1: 0.3300000

Case #2: 0.4781370

Case #3: 0.6250000

Case #4: 0.3164062

每只麻球相互独立，求一只就可以了
f[i]表示i天后1只麻球及后代全死亡的概率
f[i]=p[0]+p[1]*f[i-1]+p[2]*f[i-1]^2+........
边界f[0]=0

//

//  main.cpp

//  uva11021

//

//  Created by Candy on 26/10/2016.

//  Copyright © 2016 Candy. All rights reserved.

//

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

using namespace std;

const int N=;

int T,n,m,k;

double f[N],p[N];

inline double fp(double a,int b){

    double ans=1.0;

    for(;b;b>>=,a*=a)

        if(b&) ans*=a;

    return ans;

}

void dp(){

    f[]=;

    for(int i=;i<=m;i++){

        f[i]=p[];

        for(int j=;j<n;j++) f[i]+=p[j]*fp(f[i-],j);

    }

}

int main(int argc, const char * argv[]) {

    scanf("%d",&T);

    for(int cas=;cas<=T;cas++){

        scanf("%d%d%d",&n,&k,&m);

        for(int i=;i<n;i++) scanf("%lf",&p[i]);

        dp();

        printf("Case #%d: %.7lf\n",cas,fp(f[m],k));

    }

    return ;

}