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 <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <cmath>
#include <iomanip>
#include <map>
#include <deque>
#include <string.h>
#include <set>
using namespace std;

#define REP(i,n) for(int i = 0; i < n; i++)
#define ALL(v) v.begin(),v.end()
typedef long long ll;
typedef pair<int,int> pi;
typedef pair<ll,ll> pll;
typedef pair<string,string> pss;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pi> vpi;
typedef vector<ll> vll;
typedef vector<vll> vvll;

double EPS = 1e-9;
int INFi = 1000000005;
long long INFll = 1000000000000000005ll;
double PI = acos(-1);
int dirx[8] = {-1,0,0,1,-1,-1,1,1};
int diry[8] = {0,1,-1,0,-1,1,-1,1};
ll MOD = 1000000007;

int a, b, c, x, y;

int main(){
  cin >> a >> b >> c >> x >> y;
  if((a + b) / 2 > c) {
    if(x > y) {
      if(a > 2 * c) {
        cout << 2 * c * x << endl;
      } else {
        cout << 2 * c * y  + a * (x - y) << endl;
      }
    } else {
      if(b > 2 * c) {
        cout << 2 * c * y << endl;
      } else {
        cout << 2 * c * x + b * (y - x) << endl;
      }
    }
  } else {
    cout << x * a + y * b << endl;
  }

  return 0;
}