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>
//登録しよう！→ https://www.youtube.com/channel/UCRXsI3FL_kvaVL9zoolBfbQ
using namespace std;
using ll = long long;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}//a,bの最大公約数(gcd)を求める
ll lcm(ll a,ll b){return abs(a*b)/gcd(a,b);}//a,bの最小公倍数(lcm)を求める
vector<ll> enum_div(ll n){vector<ll> ret;for(int i=1 ; i*i<=n ; i++){if(n%i == 0){ret.push_back(i);if(i!=1 && i*i!=n)ret.push_back(n/i);}}ret.push_back(n);return ret;}
//↑nの約数を求める
vector<bool> IsPrime; void sieve(size_t max){if(max+1 > IsPrime.size())IsPrime.resize(max+1,true);IsPrime[0] = false;IsPrime[1] = false;for(size_t i=2; i*i<=max; ++i)if(IsPrime[i])for(size_t j=2; i*j<=max; ++j)IsPrime[i*j] = false;}
//↑エラトステネスの篩で素数を求める
#define roundup(divisor,dividend) (divisor + (dividend - 1)) / dividend //切り上げ割り算
#define all(x) (x).begin(),(x).end() //xの初めから終わりまでのポインタ
#define size_t ll //size_tは自動でllに変換される
#define pb(x) push_back(x)
#define gre greater //sortを降順にする
#define pri_queue priority_queue //優先度付きキュー
#define syo(x) fixed << setprecision(x) //iostreamで小数をx桁表示

void solve(long long A, long long B, long long C, long long X, long long Y){
  ll nc = C*2 , x = A*X + B*Y , ans = x;
  ll nx = X , ny = Y;
  for (int i = 0; i < max(X,Y); i++) {
    x -= ((nx > 0 ? A : 0) + (ny > 0 ? B : 0));
    if(nx > 0)nx--;
    if(ny > 0)ny--;
    x += nc;
    ans = min(x,ans);
  }
  cout << ans << endl;
}

// Generated by 1.1.6 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
int main(){
    long long A;
    scanf("%lld",&A);
    long long B;
    scanf("%lld",&B);
    long long C;
    scanf("%lld",&C);
    long long X;
    scanf("%lld",&X);
    long long Y;
    scanf("%lld",&Y);
    solve(A, B, C, X, Y);
}