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<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<map>
#include<cstdio>
#include<stack>
#include<queue>
#include<fstream>
#include<cstdio>
#include<deque>
#include<iomanip>
#include <random>

using namespace std;
#define ok1 printf("ok1\n");
#define ok2 printf("ok2\n");
#define M 1000000000000000000LL
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,s,n) for(int i=(s);i<(n);i++)
#define repr(i,n) for(int i=n-1;i>=0;i--)
#define REPR(i,s,n) for(int i=(s);i>=(n);(i)--)
#define all(a) (a).begin(),(a).end()
#define reall(a) (a).rbegin(),(a).rend()
#define pb push_back
#define pf push_front
#define MIN(a,b) a=min((a),(b))
#define MAX(a,b) a=max((a),(b))
#define SIZE(v) (int)v.size()
const double pi = acos(-1.0);

typedef vector<int> vi;
typedef vector<string> vs;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef deque<ll> dll;
ll mod = 1e9 + 7;

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

	ll a, b, c, x, y;
	cin >> a >> b >> c >> x >> y;
	ll ax = (a*x + b * y);
	ll ay = 0, by = 0;;
	if(x > y) {
		rep(i, y*2) {
			ay += c;
			//cout << ay << endl;
		}
		ll bx = x;
		x -= y;
		ay += (a*x);
		bx *= 2;
		by = c * bx;
		//cout << by << x;
		MIN(ay, by);
	}
	else {
		rep(i, x*2) {
			ay += c;
			//cout << ay << endl;
		}
		ll bx = y;
		y -= x;
		ay += (b*y);
		bx *= 2;
		by = (c*bx);
		//cout << by;
		MIN(ay, by);
	}
	//cout << ax << " " << ay << endl;
	cout << min(ax, ay) << endl;
	return 0;
}