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<algorithm>
#include<cstdio>
#include<functional>
#include<iostream>
#include<cfloat>
#include<climits>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<map>
#include<queue>
#include<set>
#include<sstream>
#include<stack>
#include<string>
#include<time.h>
#include<vector>
using namespace std;
const int oo = 0x3f3f3f3f;
int a , b , c , x , y;
int main(){
	cin >> a >> b >> c >> x >> y;
	int ans = 2 * oo;
	for(int i = 0 ; i <= max(2 * x , 2 * y) ; i += 2){
		int la = max(0 , x - i / 2);
		int lb = max(0 , y - i / 2);
		ans = min(ans , la * a + lb * b + i * c);
	} 
	cout << ans << endl;
	system("pause");
	return 0;
}