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;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define ALL(a)  (a).begin(),(a).end()
#define SUM(a)  accumulate((a).begin(),(a).end(),0)
#define SORT(a) sort(ALL(a))
#define REV(a) reverse(ALL(a))
#define dump(x)  cerr << #x << " = " << (x) << endl;
typedef long long ll;
const int INF = 999999999;
const int MOD = (int)1e9 + 7;
const int EPS = 1e-9;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int a,b,c,x,y;
  cin >> a >> b >> c >> x >> y;
  int ans = 0;
  if (2*c<=(x>y?a:b)){
      ans = 2*c*max(x,y);
  }else if (a+b>2*c){
      int mn = min(x,y);
      x -= mn;
      y -= mn;
      ans = mn*2*c + x*a + y*b;
  }else{
      ans = x*a+y*b;
  }
  cout << ans << endl;



}