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 PyPy3 (2.4.0) (judge verdict: Accepted)

import bisect
import math
import sys
from collections import Counter, defaultdict, deque
from copy import copy, deepcopy
from heapq import heapify, heappop, heappush
from itertools import combinations, permutations
from queue import Queue

read = sys.stdin.read
readline = sys.stdin.readline 
readlines = sys.stdin.readlines 

inf = float("inf")


def main():
    A, B, C, X, Y = map(int, input().split())
    if A + B > 2 * C:
        if X > Y:
            ret = 2 * C * Y
            if A > 2 * C:
                ret += 2 * C * (X - Y)
            else:
                ret += A * (X - Y)
        else:
            ret = 2 * C * X
            if B > 2 * C:
                ret += 2 * C * (Y - X)
            else:
                ret += B * (Y - X)
    else:
        ret = A * X + B * Y
    print(ret)
        
      
if __name__=='__main__':
    main()