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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>



int descending_compare(const void *a, const void *b){
    if (*(int*)a > *(int*)b){
        return -1;
    }else if (*(int*)a == *(int*)b){
        return 0;
    }else{
        return 1;
    }
}

int ascending_compare(const void *a, const void *b){
    if (*(int*)a < *(int*)b){
        return -1;
    }else if (*(int*)a == *(int*)b){
        return 0;
    }else{
        return 1;
    }
}


int lower_bound(int *a, int n, int key){
    int left, mid, right;
    left = 0, right = n;
    mid = (left + right)/2;
    while ((left+1 != mid || mid+1 != right) && mid != left){
        if (key > a[mid]){
            left = mid;
        }else{
            right = mid+1;
        }
        mid = (left + right)/2;
    }
    if (a[left] >= key)return left;
    if (a[mid] >= key)return mid;
    if (a[right] >= key)return right;
    return n;
}

unsigned long  gcd(unsigned long x, unsigned long y)
{
        if (y == 0) {
                return x;
        } else {
                return gcd(y, x % y);
        }
}

long long factorial(int x){
    long long rtn = 1;
    int i;
    for (i = x; i > 1; i--){
        rtn = (rtn*i);
    }
    return rtn;
}




int main(void){
    int price[4];
    int n;
    unsigned long long min = price[0]*(int)pow(2, 3);
    int mini;
    unsigned long long ans = 0;
    for (int i = 0; i < 4; i++){
        scanf("%d", &price[i]);
    }
    scanf("%d", &n);
    n *= 4;
    for (int i = 0; i < 3; i++){
        if (min > price[i]*(int)pow(2, 3-i)){
            min = price[i]*(int)pow(2, 3-i);
            mini = i;
        }
    }
    if (min > price[3]){
        ans = (unsigned long long)n/8*price[3]+(n%8)/(int)pow(2, mini)*price[mini];
    }else{
        ans = (unsigned long long)n/(int)pow(2, mini)*price[mini];
    }

    printf("%llu\n", ans);

    return 0;
}