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>
#include <vector>
#include <math.h>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <stack>
#include <istream>
#include <map>
#include <set>
#include <queue>
#define rep0(i,n) for (int i = 0; i < (n); ++i)  //0スタート
#define rep1(i,n) for (int i = 1; i < (n); ++i)  //1スタート
#define REP(i,a,b) for (int i = a; i < (b); ++i)  //aからbまで
#define P(x) cout<<x<<endl  //変数xを改行付きで出力
#define INF 1000000007  //最大値兼MOD
typedef long long ll;
using namespace std;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
/*
./a.exe
*/
int a,b,c,x,y;
cin>>a>>b>>c>>x>>y;
int temp = min(x,y);
int ans = temp*min(a+b,c*2);
if(x < y){
	ans += abs(x-y) * min(b,c * 2);
}
else{
	ans += abs(x-y) * min(a,c*2);
}
cout<<ans<<endl;
return 0;
}