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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

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

int main(){

	long int i,j,N,cnt=0;
	char s[110];
	long int A,B,C,X,Y,ans=0,zan;
	
	scanf("%ld%ld%ld%ld%ld",&A,&B,&C,&X,&Y);
	
	if(A>C){
		ans += 2*C*X;
		zan = Y-X;
		if(zan>0){
			if(B>C){
				ans += 2*C*zan;
			}
			else{
				ans += zan*B;
			}
		}
	}
	else if(B>C) {
		
		ans += 2*C*Y;
		zan = X-Y;
		if(zan>0){
			if(A>C){
				ans += 2*C*zan;
			}
			else ans += zan*A;
		}
		
	}
	else{
		ans += A*X+B*Y;
		
	}
	
	printf("%ld\n",ans);
	

 return 0;
 }