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 Python (3.4.3) (judge verdict: Accepted)

#!/usr/bin/env python
# coding: utf-8

# In[29]:


# Q, H, S, D = map(int, input().split())
# N = int(input())


# In[30]:


# volume = [0.25, 0.5, 1, 2]
# unit_c = [Q*4, H*2, S, D/2]
# act_c = [Q, H, S, D]

# mylist = [[volume[i], unit_c[i], act_c[i]] for i in range(len(volume))]
# mylist_sorted = sorted(mylist, key=lambda x: x[1])

# ans = 0
# target = N
# for v,c,a in mylist_sorted:
#     count = target//v
#     ans += count*a
#     target -= v*count 
# #     print(v, c, a, count, target)
#     if target == 0:
#         break
# print(int(ans))


# In[ ]:


q, h, s, d = map(int, input().split())
n = int(input())
 
cost_per_2l = min(q*8, h*4, s*2, d)
cost_per_1l = min(q*4, h*2, s)
 
a = n // 2
b = n % 2
 
print(a*cost_per_2l + b*cost_per_1l)