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: Accepted)

#include<bits/stdc++.h>
using namespace std;

#define IO ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define m(a,b) memset(a,b,sizeof(a))
#define rep(i,n) for(int i=1;i<=n;i++)
typedef long long ll;
typedef pair<string,int> p;
const int MAXN=1e5+10;
int a[105];
int main(){
	IO;
	int a,b,c,x,y;
	cin>>a>>b>>c>>x>>y;
	if (x==y) cout<<min(x*(a+b),2*x*c)<<endl;
	else{
		int ans=0;
		if (x>y){
			ans+=(x-y)*a;
			ans+=min(y*(a+b),2*y*c);
		}
		else{
			ans+=(y-x)*b;
			ans+=min(x*(a+b),2*x*c);
		}
		if (max(x,y)*2*c<ans) ans=y*2*c;
		cout<<ans<<endl;
	}
	return 0;
}