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 <iostream>
#include <array>
#include <algorithm>
#include <vector>
#include <bitset>
#include <set>
#include <unordered_set>
#include <cmath>
#include <complex>
#include <deque>
#include <iterator>
#include <numeric>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <string>
#include <tuple>
#include <utility>
#include <limits>
#include <iomanip>
#include <functional>
#include <cassert>
using namespace std;

using ll=long long;
template<class T> using V = vector<T>;
template<class T, class U> using P = pair<T, U>;
using vll = V<ll>;
using vvll = V<vll>;
#define rep(i, k, n) for (ll i=k; i<(ll)n; ++i)
#define REP(i, n) rep(i, 0, n)
#define ALL(v) v.begin(),v.end()
template < class T > inline bool chmax(T& a, T b) {if (a < b) { a=b; return true; } return false; }
template < class T > inline bool chmin(T& a, T b) {if (a > b) { a=b; return true; } return false; }
#define DEBUG_VLL(vec) REP(sz, vec.size()) std::cerr<<vec[sz]<<(sz==vec.size()-1?'\n':' ');

const long long MOD = 1000000007;
const long long HIGHINF = (long long)1e18;
const int INF = (int)1e9;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll a, b, c, x, y; cin >> a >> b >> c >> x >> y;
    ll ans = HIGHINF;
    for (int i = 0; i <= 2 * max(x + (x & 1), y + (y & 1)); i += 2) {
        ll tmp = i * c;
        ll rema = x - i / 2, remb = y - i / 2;
        tmp += a * max(rema, 0LL);
        tmp += b * max(remb, 0LL);
        chmin(ans, tmp);
    }
    cout << ans << '\n';
    return 0;
}