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<map>
#include<cstring>
#include<math.h>
#include<bitset>
#include<queue>
#include<set>
#include<iomanip>
#include<math.h>
#include<assert.h>
#include<string>
#include<cstdlib>
#include<cctype>
// #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
constexpr ll INF = 1LL << 40;
const double INFD = 1e100;
const ll mod = 1000000007;
const double PI = 3.1415926535897932384626433832795028841971;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
// ----------------------------------------------------------------------------

int main(){
  int A,B,C,X,Y;
  cin >> A >> B >> C >> X >> Y;
  int ans=mod;
  for(int c=0; c<=200000; c+=2){
    int sum=0;
    if(X>c/2){
      sum += (X-c/2)*A;
    }
    if(Y>c/2){
      sum += (Y-c/2)*B;
    }
    sum += c*C;
    chmin(ans,sum);
  }
  cout << ans << endl;
  return 0;
}