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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
long long A,B,C,X,Y,p,ans;
int main()
{
	scanf("%lld%lld%lld%lld%lld",&A,&B,&C,&X,&Y);X*=2;Y*=2;
	if(2*C<A+B){
		ans+=min(X,Y)*C;
		if(X<=Y){
			Y-=X;
			if(2*C<B)ans+=Y*C;
			else ans+=(Y/2)*B;
		}
		else{
			X-=Y;
			if(2*C<A)ans+=X*C;
			else ans+=A*(X/2);
	}
}
	else{
		int p=min(X,Y);
		ans+=B*(p/2);
		ans+=A*(p/2);X-=p;Y-=p;
		//cout<<ans<<endl;
	if(X)
		if(2*C>A)ans+=A;
	else ans+=C*2;
	else
		if(2*C>B)ans+=B;
	else ans+=C*2;
	}
	cout<<ans<<endl;
	return 0;
}