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;
using ll =long long;
#define SORT(a) sort((a).begin(),(a).end())
#define rSORT(a) reverse((a).begin(),(a).end())
#define For(i, a, b)    for(int i = (a) ; i < (b) ; ++i)
#define rep(i, n)       For(i, 0, n)
#define debug(x)  cerr << #x << " = " << (x) << endl;
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
void coY() {cout <<"Yes"<<endl;}
void coN(){cout <<"No"<<endl;}
const ll INF = 1LL << 60;
const int mod = 1e9+7;
//Write From this Line

int main()
{
	int a , b,  c, x,  y;
	cin >> a >> b>>  c>> x>>  y;
	ll ans = INF; ;
	for(int i = 0 ; i <=  max(x,y) ; i++ ){
		//2i個のcを買う
		ll tmp = 0;
		tmp += 2*c * i ;
		tmp += max(0,x-i) * a  + max(0,y-i) * b;
		chmin(ans,tmp);
	}
	cout << ans << endl;
}