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 <ios>
#include <iomanip>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <utility>
using namespace std;

#define MOD 1000000007
#define INF 1e9+7
#define INF_LONG 1e18
#define rep(i, n) for(int i=0; i<n; ++i)
#define rrep(i, m, n) for(int i=m; i<n; ++i)
typedef long long ll;

#define MAX_N 200000
#define MAX_M 100000
#define MAX_V 100
#define MAX_E 100
#define MAX_Q 100000

int dx[] = {-1, 0, 1, 0};
int dy[] = {0, -1, 0, 1};

int ddx[] = {-1, 0, 1, 0, -1, 1, -1, 1};
int ddy[] = {0, -1, 0, 1, 1, 1, -1, -1};

ll gcd(ll x, ll y) {
  if(y==0) return x;
  else return gcd(y, x%y);
}

ll lcm(ll x, ll y) {
  return (x*y)/gcd(x, y);
}

//--------------------------

int main() {

  int a, b, c; cin >> a >> b >> c;
  int x, y; cin >> x >> y;
  ll cost = 0;

  int slice_min = min(x, y);

  if(a+b>2*c) {
    cost += 2*c*slice_min;
  }else {
    cost += a*slice_min + b*slice_min;
  }

  if(x-slice_min!=0) {
    if(a>2*c) cost += 2*c*(x-slice_min);
    else cost += a*(x-slice_min);
  }else {
    if(b>2*c) cost += 2*c*(y-slice_min);
    else cost += b*(y-slice_min);
  }

  cout << cost << endl;

  return 0;
}