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: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,n)for(long long i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

const int MOD=1000000007;
const int INF=0x3f3f3f3f;
const ll INFL=0x3f3f3f3f3f3f3f3f;
ll gcd(ll x, ll y){
  if(y==0) return x;
  else return gcd(y, x%y);
}
ll lcm(ll x,ll y){
  ll g=gcd(x,y);
  return x/g*y;
}
 
int main() {
  int A, B, C, X, Y;
  cin >> A >> B >> C >> X >> Y;
  
  int ans;
  if(A+B<2*C){
    ans=A*X+B*Y;
  }
  else{
    ans=min(X,Y)*C*2;
    int Z=min(X,Y);
    X-=Z;
    Y-=Z;
    if(Y==0){
      if(A<2*C) ans+=A*X;
      else ans+=X*C*2;
    }
    if(X==0){
      if(B<2*C) ans+=B*X;
      else ans+=Y*C*2;
    }
  }
  cout<<ans<<endl;
}