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 <math.h>
#include <numeric>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <queue>
#include <queue>
#include <sstream>
#include <set>
typedef  long long ll;
const int dx[8]={1,0,-1,0,1,-1,-1,1};
const int dy[8]={0,1,0,-1,1,1,-1,-1};
const int dr[4] = {-1, 0, 1, 0};
const int dc[4] = { 0, 1, 0,-1};
const int INF = 1e9;
#define FOR(i, a, n) for(int i = (int)(a); i < (int)(n); ++i)
#define REP(i, n) FOR(i, 0, n)
#define SORT(a) sort(a.begin(),a.end())
#define REVERSE(a) reverse(a.begin(),a.end())
int guki (int a){
    if(a%2==0) return 0;
    else return 1;
}
int gcd(int a, int b){
    if(a%b==0){
        return b;
    }else{
        return (gcd(b,a%b));
    }
}
int lcm(int a, int b){
    int x = gcd(a,b);
    return (a*b/x);
}



using namespace std;
int main(){
    int a,b,ab,x,y;
    cin >> a >> b >> ab >> x >>y;

    int  k = max(x,y);

    int ans = INF;

    for (int i = 0; i<=k; i++){
        int q = max(0, x-i);
        int w = max(0,y-i);

        ans = min(ans, a*q + b*w + i*ab*2 );
    }

    cout << ans <<endl;
}