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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<string>
#include<map>
//#define ll long long
using namespace std;
const int maxn=100010;
int v[maxn];
int plate[maxn];

int main( )
{
    int a,b,c,x,y;
    long long sum;
    cin>>a>>b>>c>>x>>y;//一盒南方规则棋盘价格为A，一盒北方规则棋盘价格为B
                      //X盒适用南方规则的棋盘和Y盒北方规则的棋盘
    int k=min(x,y);
    if(2*c<a+b)
        sum=c*k*2;
    else
        sum=(a+b)*k;
    if(x>k)
    {
        sum+=a*(x-k);
    }
    if(y>k)
    {
        sum+=b*(y-k);
    }
    cout<<sum;
	 return 0;
}