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;

// マクロ
// forループ
#define REP(i,n) for(ll i=0;i<(ll)(n);i++)
#define REPD(i,n) for(ll i=n-1;i>=0;i--)
#define FOR(i,a,b) for(ll i=(a);i<=(ll)(b);i++)
#define FORD(i,a,b) for(ll i=(a);i>=(ll)(b);i--)
// xにはvectorなどのコンテナ
#define ALL(x) (x)begin(),(x).end()
#define SIZE(x) ((ll)(x).size())
#define MAX(x) *max_element(ALL(x))
#define MIN(x) *min_element(ALL(x))
// 出力
#define PRINT(x) cout<<(x)<<endl
#define PRINTA(x,n) REP(i,n)PRINT(x[i])
// 定数
#define INF 1000000000000 // 10^12
#define MOD 1000000007 // 10^9+7
// 略記
#define PB push_back
#define MP make_pair

int A, B, C, X, Y;

int main() {
    cin >> A >> B >> C >> X >> Y;
    int min_cost = 0;
    if (A + B < 2 * C) {
        min_cost += A * X + B * Y;
    }
    else if (X < Y) {
        min_cost += 2 * C * X;
        if (B < 2 * C) {
            min_cost += B * (Y - X);
        }
        else {
            min_cost += 2 * C * (Y - X);
        }
    }
    else {
        min_cost += 2 * C * Y;
        if (A < 2 * C) {
            min_cost += A * (X - Y);
        }
        else {
            min_cost += 2 * C * (X - Y);
        }
    }
    PRINT(min_cost);
    return 0;
}