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)

//============================================================================
//Everyone has a different way of thinking, so God Created us
//   Hope You Respect My Way..,Thank You ):
// Author      : Murad
// Online Judge: Codeforces.cpp & Atcoder.cpp
// Description : Problem name
//============================================================================
#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
#include<utility>
using namespace std;
typedef long long LL;
#define pi 3.1415926536
#define forn(i,a,b)for(int i=a;i<b;i++)
#define ULL unsigned long long
#define MP make_pair
#define ff first
#define ss second
#define endl '\n'
#define INF LL(1e18);
#define _INF INT32_MIN
#define pq priority_queue
#define MM multimap
#define PB push_back
#define EMPP emplace_back
#define vii vector<int>
#define vll vector<LL>
#define ipair pair<int,int>
#define lpair pair<LL,LL>
#define El3zba ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define modd 1000000007
#define sf1(v) scanf("%I64d",&v);
#define sf2(v1,v2) scanf("%I64d %I64d",&v1,&v2)
#define sf3(v1,v2,v3) scanf("%I64d %I64d %I64d",&v1,&v2,&v3)
//std::transform(s1.begin(), s1.end(), s1.begin(),::tolower);
char alphz[27] = { 'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z' };
float Euclidean(LL x1, LL x2, LL y1, LL y2) {
	return sqrt(pow(x1 - x2, 2) + pow(y1 - y2, 2));
}
LL GCD(LL a, LL b) {
	return !b ? a : GCD(b, a % b);
}
LL LCM(LL a, LL b) {
	return (a*b) / GCD(a, b);
}
void PrimeFactor(LL n)
{
	while (n % 2 == 0) {
		//printf("%d ", 2);
//		aa.insert(2);
		n /= 2;
	}
	for (int i = 3; i <= sqrt(n); i += 2)
	{
		if (n%i == 0) {
			//printf("%d ", i);
			//aa.insert(i);
			n /= i;
		}
	}
	if (n > 2) {
		//printf("%d ", n);
	//	aa.insert(n);
	};
}
bool Is_Square(LL x) {
	LL l = 0, r = x;
	while (l <= r) {
		LL mid = l + (r - l) / 2;
		if (mid*mid == x)return true;
		if (mid*mid > x)r = mid - 1;
		else l = mid + 1;
	}
	return false;
}
LL Power(LL x, LL y)
{
	LL temp;
	if (y == 0)return 1;
	temp = Power(x, y / 2);
	if (y % 2 == 0)
		return temp*temp;
	else return x*temp*temp;
}
bool Is_Prime(int x) {
	if (x == 2)return 1;
	else if (x % 2 == 0 || x < 2)return 0;
	for (int i = 3; i*i <= x; i += 2)
		if (x%i == 0)return 0;
	return 1;
}
LL solv(int a, int b, int c, int x, int y)
{
	if (x <= 0 && y <= 0)return 0;
	LL go1 = 0;
	if (x > 0 && y > 0)
	{
		go1=min(solv(a, b, c, x - 1, y - 1) +( a + b), solv(a, b, c, x - 1, y - 1) + 2 * c);
	}
	else if (x > 0 && y <= 0)
	{
		go1+= x*a;
	}
	else if(x<=0&&y>0)
		go1 +=y*b;
	return go1;
}
int main()
{
	El3zba;
	/***  بسم الله الرحمن الرحيم   ***/
	int a, b, c, x, y;
	cin >> a >> b >> c >> x >> y;
	LL ans = solv(a, b, c, x, y);
	cout << ans << endl;
	return 0;
}