"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.

Examples

Input

1500 2000 1600 3 2


Output

7900


Input

1500 2000 1900 3 2


Output

8500


Input

1500 2000 500 90000 100000


Output

100000000
A, B, C, X, Y = map(lambda x: int(x), raw_input().split())


"""                                                                                                                                                                   
                                                                                                                                                                      
1500 2000 1600 3 2                                                                                                                                                    
                                                                                                                                                                      
A * 3 + B * 2                         = 8500                                                                                                                          
                                                                                                                                                                      
2C + A*2 + B * 1 = 3200 + 4000 + 1600 = 8800                                                                                                                          
4C + A*1 + B * 0 = 6400 + 1500        = 7900                                                                                                                          
6C                                    = 9600                                                                                                                          
                                                                                                                                                                      
"""

acc = []

maxC = 2 * max(X,Y) + 2

for i in range(maxC):
    x=i
    if x % 2 == 0:
        Csize = x/2
        remXsize = X - Csize
        remYsize = Y - Csize
        acc.append( C * x + A * max(remXsize,0) + B * max(remYsize,0) )

#print acc                                                                                                                                                            
print min(acc)