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 <algorithm>  //min, max, sort, unique, fill
#include <iostream>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>  //pair, swap,
#include <vector>
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
using VI = vector<ll>;

#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define REP(i, n) for (ll i = 0; i < (n); i++)

static const ll kIntMax = numeric_limits<ll>::infinity();
static const double kFloatMax = numeric_limits<double>::infinity();

vector<ll> vi;

int main() {
  ll a, b, c, x, y;
  cin >> a >> b >> c >> x >> y;
  ll price = 0, tmp;
  // price = 1000000000;
  // REP(i, 100001) {
  //   tmp = i * c * 2 + max(0ll, x - i) * a + max(0ll, y - i) * b;
  //   if (tmp < price) price = tmp;
  //   if (i > x && i > y) break;
  // }

  if (a + b < c * 2)
    price = a * x + b * y;
  else {
    if (c * 2 <= a) {
      price += x * c * 2;
      y -= x;
      x = 0;
      // cerr << "price:" << price << "x:" << x << "y:" << y << endl;
    }

    if (c * 2 <= b && y > 0) {
      price += y * c * 2;
      x -= y;
      y = 0;
      // cerr << "price:" << price << "x:" << x << "y:" << y << endl;
    }

    x = max(0ll, x);
    y = max(0ll, y);

    ll z = min(x, y);
    price += z * c * 2 + (x - z) * a + (y - z) * b;
    // cerr << "price:" << price << "x:" << x << "y:" << y << "z:" << z <<endl;
  }

  cout << price << endl;

  return 0;
}