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)

//  cout << char(i+48) << endl;//なぜかaは47と得る時がある。+48で出力もaにできる。
//  cout << char(97) << endl;//アスキーコードでaを出力
// sort(b.begin(), b.end());//bという配列を小さい方からソート
// reverse(b.begin(), b.end());//bという配列をリターン
// 10桁でdouble出力するパターン１ 
// printf("%.10f\n", ma);
// 10桁でdouble出力するパターン２
//cout << fixed << setprecision(10);
//  cout << ma << endl;

/*01 02 03 12 13 23　と６回見ていくパターン 
for(int i=0;i<n-1;i++){
  for(int j=i+1;j<n;j++){
    }
  }
*///vector<vector<int>> data(3, vector<int>(4));//int型の2次元配列(3×4要素の)の宣言
//int64_t a;
//10のi乗pow(10, i);
/*string s; stringでの文字列を数字型に変える方法
  cin >> s;
  rep(i,s.size()-2) {
  int a= (s.at(i)-'0')*100 + (s.at(i+1)-'0')*10+ s.at(i+2) -'0';
    */
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
int main() {
  int a,b,c,x,y;
  cin >> a >> b >> c >> x >> y;
  int A,B,C;
  A = a * x + b * y;
  if(x>y) B = c*y*2+a*(x-y);
  else    B = c*x*2+b*(y-x);
  
  if(x>y) C = c*x*2;
  else    C = c*y*2;
  int ma = A;
  ma = min(ma,A);
  ma = min(ma,B);
  ma = min(ma,C);
  cout << ma << endl;
//  cout << A << B << C << endl;
}