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 (Clang 3.8.0) (judge verdict: Accepted)

#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>
#include <climits>

#define rep(i, m, n) for(int i=int(m);i<int(n);i++)
#define EACH(i, c) for (auto &(i): c)
#define all(c) begin(c),end(c)
#define EXIST(s, e) ((s).find(e)!=(s).end())
#define SORT(c) sort(begin(c),end(c))
#define pb emplace_back
#define MP make_pair
#define SZ(a) int((a).size())

//#define LOCAL 0
//#ifdef LOCAL
//#define DEBUG(s) cout << (s) << endl
//#define dump(x)  cerr << #x << " = " << (x) << endl
//#define BR cout << endl;
//#else
//#define DEBUG(s) do{}while(0)
//#define dump(x) do{}while(0)
//#define BR
//#endif


//改造
typedef long long int ll;
using namespace std;
#define INF (1 << 30) - 1
#define INFl (ll)5e15
#define DEBUG 0 //デバッグする時1にしてね
#define dump(x)  cerr << #x << " = " << (x) << endl
#define MOD 1000000007
//ここから編集する


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

    ll A, B, C, X, Y;
    cin >> A >> B >> C >> X >> Y;

    ll ans = 0;

    if (2 * C > A + B) {
        ans = A * X + B * Y;
    } else {
        ll Z = min(X, Y);
        ans += 2 * Z * C;
        X -= Z;
        Y -= Z;

        if (X == 0) {
            if (B <= 2 * C) {
                ans += B * Y;
            } else {
                ans += Y * C * 2;
            }
        } else{
            if (A <= 2 * C){
                ans += A * X;
            }else{
                ans += X * C * 2;
            }
        }
    }
    cout << ans << endl;


    return 0;
}