Problem:
Score : 300 points

Problem Statement

You've come to your favorite store Infinitesco to buy some ice tea.

The store sells ice tea in bottles of different volumes at different costs.
Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen.
The store has an infinite supply of bottles of each type.

You want to buy exactly N liters of ice tea. How many yen do you have to spend?

Constraints

1 \leq Q, H, S, D \leq 10^8

1 \leq N \leq 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

Q H S D
N

Output

Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.

Sample Input 1

20 30 70 90
3

Sample Output 1

150

Buy one 2-liter bottle and two 0.5-liter bottles. You'll get 3 liters for 90 + 30 + 30 = 150 yen.

Sample Input 2

10000 1000 100 10
1

Sample Output 2

100

Even though a 2-liter bottle costs just 10 yen, you need only 1 liter.
Thus, you have to buy a 1-liter bottle for 100 yen.

Sample Input 3

10 100 1000 10000
1

Sample Output 3

40

Now it's better to buy four 0.25-liter bottles for 10 + 10 + 10 + 10 = 40 yen.

Sample Input 4

12345678 87654321 12345678 87654321
123456789

Sample Output 4

1524157763907942

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <bitset>
# include <sstream>
# include <set>
# include <cmath>
# include <algorithm>
# pragma  comment(linker,"/STACK:102400000,102400000")
using namespace std;
# define LL          long long
# define pr          pair
# define mkp         make_pair
# define lowbit(x)   ((x)&(-x))
# define PI          acos(-1.0)
# define INF         0x3f3f3f3f3f3f3f3f
# define eps         1e-8
# define MOD         1000000007

inline int scan() {
    int x=0,f=1; char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
inline void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
# define MX 100005
/**************************/
LL q,h,s,d;
LL get(LL n)
{
    LL ans;
    if (n==1)
    {
        ans=min(4*q,2*h);
        ans=min(ans,s);
    }
    else if (n==2)
    {
        LL ppp = min(8*q,4*h);
        LL kkk = min(2*s,d);
        ans=min(ppp,kkk);
    }
    else
    {
        if (n%2)
        {
            LL p = n/2*2;
            LL aa=p*q*4;
            LL bb=p*h*2;
            LL cc=p*s;
            LL dd=p*d/2;
            ans = min(aa,bb);
            ans = min(cc,ans);
            ans = min(dd,ans);

            LL sb=min(4*q,2*h);
            sb=min(sb,s);

            ans+=sb;
        }
        else
        {
            LL p = n/2*2;
            LL aa=p*q*4;
            LL bb=p*h*2;
            LL cc=p*s;
            LL dd=p*d/2;
            ans = min(aa,bb);
            ans = min(cc,ans);
            ans = min(dd,ans);
        }
    }
    return ans;
}

int main()
{

    LL n;
    scanf("%lld%lld%lld%lld",&q,&h,&s,&d);
    scanf("%lld",&n);
    LL ans=get(n);
    printf("%lld\n",ans);
    return 0;
}