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)

//bad_bat
#include<bits/stdc++.h>
using namespace std;

#define Fast ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0)
#define FO freopen("in.txt", "r", stdin)
#define FC freopen("out.txt", "w", stdout)
#define aise cout<<"aise"<<endl
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ppf pop_front
#define mp make_pair
#define ll long long
#define all(x) x.begin(), x.end()
#define mset(v, a) memset(v, a, sizeof(v))
#define pll pair< ll, ll >
#define pdd pair< double, double >
#define ff first
#define ss second
#define pi acos(-1.0)
#define mxN 2010
#define inf 1e14
#define MOD 1000000007

ll dx[] = {0, 0, 1, -1};
ll dy[] = {1, -1, 0, 0};

int main()

{

    ll a, b, c, x, y;
    cin >> a >> b >> c >> x >> y;
    ll mx = max(x,y);
    mx*=2;
    ll cst = inf;
    for(ll i = mx; i >= 0; i--)
    {

        ll abc = i*c;
        ll acab=i/2;
        ll an=0, bn=0;
        if(x > acab) an=x-acab;
        if(y > acab) bn = y-acab;
        ll tc = abc+an*a+bn*b;
        //cout << "CH " << i << " " << acab << " " <<
        cst = min(cst,tc);
    }
    cout << cst;
}