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<bitset>


using namespace std;

#define M 1000000007
#define pb push_back
#define f first
#define s second
#define rep(i, st, ed) for(int i=st; i<ed; i++)
#define repn(i, st, ed) for(int i=st; i<=ed; i++)
#define repb(i, ed, st) for(int i=ed; i>=st; i--)

typedef long long ll;
typedef unsigned long long int llt;
typedef long double ld;

const long double PI  =3.141592653589793238463;
const int N = 1010;
const ll INF = 1LL << 60;

template<class T> void chmax(T &a,T b) { if (a<b) a=b;}
template<class T> void chmin(T &a,T b) { if (a>b) a=b;}



// Main Code
int main()
{
	ll a, b, c,x, y;
	cin>>a>>b>>c>>x>>y;
	if(2*c > (a+b))
	{
		cout<<a*x+y*b<<endl;
	}
	else
	{
		ll z =min(x, y);
		ll ans = z*c*2;
		x -= z, y -= z;
		if(x == 0)
		{
			ll l = y*c*2;
			l = min(l, y*b);
			cout<<(ans+l)<<endl;
		}
		else
		{
			ll r = x*c*2;
			r = min(r, x*a);
			cout<<(ans+r)<<endl;
		}
	}
}