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<iomanip>
#include<queue>
#include<map>
#include<bitset>
#include<math.h>
using namespace std;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define REP(i,m,n) for(int i=m;i<(int)(n);i++)
typedef pair<int,int> pint;
typedef long long ll;
const int inf   = 1e9+7;
const ll  longinf = 1LL<<60;
const ll mod=1e9+7;
int dx[4]={1,0,-1,0}, dy[4]={0,1,0,-1};

int main(){
  ll a,b,c,n,m;
  cin>>a>>b>>c>>m>>n;
  ll ret=longinf;
  rep(i,n+1){
    ll res=i*b;
    res+=2*(n-i)*c;
    ll num=max(0LL,m-(n-i));
    if(a<=2*c)res+=num*a;
    else res+=2*num*c;
    ret=min(res,ret);
    }
  cout<<ret<<endl;
  return 0;
  }