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>
using namespace std;
#include <iostream>
#include <vector>
#include <cstdlib>
#include <cmath>
#include <math.h>
#include <sstream>
#include <numeric>
#include <cctype>
#include <bitset>
#include <cassert>
#include<algorithm>
//---------------------------------------------------------------
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (int)(n); i++)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)
#define SIZE 100005
#define INF 1e18
#define all(x) x.begin(),x.end()
#define fi first
#define se second
#define vec vector
using pint = pair<int,int>;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
//---------------------------------------------------------------
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const double pi = acos(-1);
//---------------------------------------------------------------

//↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓


int main() {
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

    ll a,b,c,x,y;
    cin>>a>>b>>c>>x>>y;
    ll ans = INF;
    rep(i,100001){
        ans=min(ans,c*2*i+max(0ll,x-i)*a+max(0ll,y-i)*b);
    }

    cout<<ans<<endl;
    return 0;
}