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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
typedef long long ll;
#define ALL(l)  (l).begin(),(l).end()
#define rep(i,n) for(ll (i)=0;(i)<(n);(i)++)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)
using namespace std;

//const ll mod =  1000000007;
//const ll maxn = 1000000000;

//---------------------------------------------------------------------------------------------------
using vi = vector<int>; // intの1次元の型に vi という別名をつける
using vll = vector<ll>; // intの1次元の型に vi という別名をつける
using vvll = vector<vll>; // intの2次元の型に vvi という別名をつける
using vs = vector<string>; // stringの1次元の型に vs という別名をつける
using pll = pair<ll, ll>; // これ以降 pii という型名はpair<ll, ll> と同じ意味で使える
//---------------------------------------------------------------------------------------------------

int main() {
    ll a,b,c,x,y,ans=1000000000;
    cin>>a>>b>>c>>x>>y;
    for(ll i=0;i<=2*max(x,y);i+=2){
        ll k=i*c;
       // k+=max(0,1);
       ll zero=0;
        k+=max(zero,x-i/2)*a;
        k+=max(zero,y-i/2)*b;
        ans=min(k,ans);
    }
    cout<<ans<<endl;
    
}