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 <string.h>
#include <vector>
#include <algorithm>
#include <cmath>
#include <map>
#include <functional>
#include <cassert>
#include <utility>
#include <stack>
#include <queue>
using namespace std;
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define REP(i,n) for(int i=0;i<n;i++)
#define FOR(i,n1,n2) for(int i=n1;i<n2;i++)
#define speed_up    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
typedef pair<int,int> Pi;
typedef long long int ll;
const int INF=(ll)(1<<30)-1;
const ll INFl=(ll)9223372036854775807;
const int MAX=10000;
const ll MOD=(ll)1e9+7;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a/gcd(a, b)*b;}


int a,b,c,x,y;

int main(){
	cin>>a>>b>>c>>x>>y;
	if(a+b<=2*c){
		cout<<a*x+b*y<<endl;
		return 0;
	}
	int ans=0;
	if(x>y){
		ans+=2*c*y;
		if(a>=2*c){
			ans+=2*c*(x-y);
		}else{
			ans+=a*(x-y);
		}
	}else{
		ans+=2*c*x;
		if(b>=2*c){
			ans+=2*c*(y-x);
		}else{
			ans+=b*(y-x);
		}
	}
	cout<<ans<<endl;
	return 0;
}