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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
typedef long long ll;
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#define ios ios_base::sync_with_stdio(0) ; cin.tie(0) ; cout.tie(0)
#define so sizeof
#define pb push_back
#define cl clear() ;
#define vl vector<ll>
#define sz size()
#define len length()
#define emp empty()
#define el endl;cout.flush()
#define be begin()
#define fi first
#define se second
#define br break
#define en end()
#define ro return 0
#define eb emplace_back
#define con continue
#define ms(x) memset(x , 0, so x)
#define all(x) (x).be , (x).en
#define acc(x) accumulate((x).be , (x).en , 0ll)
#define forn(i,a,b) for(ll i=a;i<=b;++i)
#define rng_58 mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
#define vll vector<pair<ll,ll> >
#define pll pair<ll,ll>
#define plll pair<ll,pair<ll,ll>>
#define plvl pair<pair<ll,ll> ,ll>
#define mp make_pair
#define trace3(a,b,c) cerr <<"a is " << a << " b is " << b << " c is " << c << el;
#define trace4(a,b,c,d) cerr <<"a is " << a << " b is " << b << " c is " << c <<" d is " << d << el ;
#define trace5(a,b,c,d,e) cerr <<"a is " << a << " b is " << b << " c is " << c <<" d is " << d << " e is " << e << el;
#define trace6(a,b,c,d,e,f) cerr <<"a is " << a << " b is " << b << " c is " << c <<" d is " << d << " e is " << e << " f is " << f << el ;

int main()
{
    ios;
    ll a , b , c , x , y;
    cin >> a >> b >> c >> x >> y;
    ll mab = 2 * c;
    ll aab = a + b;
    if(mab <= aab)
    {
        ll foo = 0 ;
        foo = foo + mab * min(x , y);
        ll rem = max(x , y) -  min(x , y) ;
        if(x < y){
            ll ekkaniraj = min(b , 2*c);
            foo = foo + ekkaniraj * rem;
            cout << foo << el ;
            ro ;
        }
        else if(x > y){
            ll ekkaniraj = min(a , 2*c);
            foo = foo + ekkaniraj * rem ;
            cout << foo << el ;
            ro ;
        }
        else {
            cout << foo << el ;
            ro ;
        }
     }
     else if(mab > aab)
     {

         ll foo = 0 ;
         foo = aab * min(x , y) ;
         ll rem = max(x , y) - min(x , y) ;
         if(x < y){
            ll ekkaniraj = min(b , 2*c) ;
            foo = foo + ekkaniraj * rem ;
            cout << foo << el ;
            ro ;
         }
         else if(x > y ){
            ll ekkaniraj = min(a , 2*c) ;
            foo = foo + ekkaniraj * rem ;
            cout << foo << el;
            ro ;
         }
         else cout << foo << el ;
         ro ;
     }
     ro ;
}