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;

typedef long long ll;
typedef pair<ll, ll> P;
typedef vector<int> VI;
typedef vector<VI> VVI;
const double EPS = 1e-12;
const int INF = numeric_limits<int>::max() / 2;
const int NEG_INF = numeric_limits<int>::min() / 2;
const int MOD = 1e9 + 7;

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

    int a,b,c,x,y;cin>>a>>b>>c>>x>>y;
    int num=max(x,y);
    ll res=1e9;
    for(int i=0;i<=2*num;i+=2){
        ll tmp=c*i;
        if(x-i/2>0) tmp += a*(x-i/2);
        if(y-i/2>0) tmp += b*(y-i/2);
        res=min(res,tmp);
    }
    cout<<res<<endl;
}