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>
#define rep(i, n) for(int i = 0; i < n; ++i)
#define repR(i, n) for(int i = n; i >= 0; ++i)
#define FDS(i, n) for(int i = 0; i < n; ++i)
#define FDSR(i, n) for(int i = n; i >= 0; ++i)
#define FOR(i, m, n) for(int i = m; i < n; ++i)
#define FORR(i, m, n) for(int i = m;i >= n;--i)
#define VSORT(v) sort(v.begin(), v.end());
#define INF 999999999
#define itn int
#define ednl endl
using namespace std;
typedef long long ll;
int main(){
	ll A,B,C,X,Y;
	cin>>A>>B>>C>>X>>Y;
	if(X==Y){
		cout<<min((C*X*2),((A+B)*X))<<endl;
	}
	else{
	    ll sum=0;
		int t=min(X,Y);
		sum+=(min((C*2),(A+B)))*t;
		sum+=(abs(X-Y))*min(A,2*C);
		cout<<sum<<endl;
	}
}