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 Julia (0.5.0) (judge verdict: Wrong Answer)

parseInt(x) = parse(Int, x)
parseBigInt(x) = parse(BigInt, x)
parseFloat(x) = parse(Float64, x)
parseBigFloat(x) = parse(BigFloat, x)
parseMap(x::Array{SubString{String},1}) = map(parseInt, x)

function main()
    arr=parseMap(split(readline()))
    n=parseInt(readline())
    res=[(0,0) for i in 1:4]
    for i in 1:4
        tmp=2.0^(i-3)
        res[i]=(n-n÷tmp*tmp,n÷tmp*arr[i])
    end
    ans=Inf
    for (i,j) in res
        for k in 1:3
            tmp=2.0^(k-3)
            ans=min(ans,j+i÷tmp*arr[k])
        end
    end
    println(BigInt(ans))
end
main()