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: Wrong Answer)

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>

using namespace std;

#define   ll            long long
#define   PI            atan(1.0)*4
#define   mst(x,v)      memset(x,v,sizeof(x))
#define   rush()        int T;scanf("%d",&T);while(T--)
#define   debug         puts("debug")

typedef   pair<int,int> P;

const int   M = 2e5+5000;
const int mod = 1e9+7;
const int inf = 1e9;

int main() {

	ll a,b,c,x,y,ans=0;
	scanf("%lld %lld %lld %lld %lld",&a,&b,&c,&x,&y);
	
	if(2*c < a && 2*c < b) {
		ans = max(x,y)*2*c;
	}
	else if(2*c < a ) {
		ans = x*2*c + b*(max((ll)0,y - x));
	}
	else if(2*c < b ) {
		ans = y*2*c + a*(max((ll)0,x - y));
	}
	else
	{
		ans = a*x + b*y;
	}
	
	printf("%lld\n",ans);
	return 0;
}

/*
	freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
*/