Problem:
Score : 300 points

Problem Statement

"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively.

Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.

Constraints

1 ≤ A, B, C ≤ 5000

1 ≤ X, Y ≤ 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C X Y

Output

Print the minimum amount of money required to prepare X A-pizzas and Y B-pizzas.

Sample Input 1

1500 2000 1600 3 2

Sample Output 1

7900

It is optimal to buy four AB-pizzas and rearrange them into two A-pizzas and two B-pizzas, then buy additional one A-pizza.

Sample Input 2

1500 2000 1900 3 2

Sample Output 2

8500

It is optimal to directly buy three A-pizzas and two B-pizzas.

Sample Input 3

1500 2000 500 90000 100000

Sample Output 3

100000000

It is optimal to buy 200000 AB-pizzas and rearrange them into 100000 A-pizzas and 100000 B-pizzas. We will have 10000 more A-pizzas than necessary, but that is fine.

Solution in Python (3.4.3) (judge verdict: Wrong Answer)

import math
import copy
from queue import Queue
import heapq
import numpy as np

MOD = 5000000007

#n = int(input())
#b = int(input())
a, b, c, x, y = [int(i) for i in input().split()]
#table = [[int(i)  for i in input().split()] for j in range(m)]
#table = [list(input()) for n in range(r)] #table[h][w]
#p = [int(i) for i in input().split()]
#s = [int(i) for i in range(n)]
#p = [int(input()) for i in range(m)]
#s = input()

#list = list(s)
#dp = [MOD for _ in range(m)]
#dp = [[0 for _ in range(n+1)] for _ in range(n+1)]
#dp = [[[0 for _ in range(11)] for _ in range(2)] for _ in range(11)]

#Combination = [[0 for _ in range(n+1)] for _ in range(n+1)]
#b = [[0 for _ in range(n+1)] for _ in range(n+1)]

kotae = 500000001
if (x > y):
    max = x
else:
    max = y
for i in range(0,max+1):
    j = x - i
    k = y - i
    if(j < 0):
        j = 0
    if(k < 0):
        k = 0
    temp = a * j + b * k + 2 * c * i
    if (temp < kotae):
        kotae = temp

print(kotae)