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 fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(x) (x).begin (),(x).end()
#define sqrt(x) sqrt(abs(x))
#define re return
#define sz(x) ((int)(x).size ())
#define prev PREV
#define next NEXT
 
using ll = long long;
using ii = pair<int, int>;
using ld = long double;
using D = double;
using vi = vector<int>;
using vii = vector<ii>;
using vvi = vector<vi>;
using vs = vector<string>;
 
template<typename T> T abs (T x) { re x < 0 ? -x : x; }
template<typename T> T sgn (T x) { re x < 0 ? -1 : (x > 0 ? 1 : 0); }
template<typename T> T sqr (T x) { re x * x; }
template<typename T> T gcd (T a, T b) { re a ? gcd (b % a, a) : b; }
 
int n;
int m;
int a, b, ab, na, nb;
 
int main () {
    cin >> a >> b >> ab >> na >> nb;
    int ans = 1e9;
    for (int t = 0; t <= max (na, nb); t++)
        ans = min (ans, 2 * t * ab + max (0, na - t) * a + max (0, nb - t) * b);
    cout << ans << endl;
	return 0;
}