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)

#pragma region template
#include "bits/stdc++.h"
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
const long long MOD = 1e9 + 7, MAX = 1e18, larg = 1e5, INF = -1e9;
long long A, B, C, D, E, F, G, H, I, J, K, L, N, M, O, P, Q, R, S, T, U, V, W, x, y, z;
long long max_value = INF, max_index = -1;
long long min_value = MAX, min_index = -1;
typedef long long ll;
typedef pair<ll, ll> Pa;

template <std::int_fast64_t Modulus> class modint {
	using u64 = std::int_fast64_t;
public:
	u64 a;
	constexpr modint(const u64 x = 0) noexcept : a(x% Modulus) {}
	constexpr u64& value() noexcept { return a; }
	constexpr const u64& value() const noexcept { return a; }
	constexpr modint operator+(const modint rhs) const noexcept {
		return modint(*this) += rhs;
	}
	constexpr modint operator-(const modint rhs) const noexcept {
		return modint(*this) -= rhs;
	}
	constexpr modint operator*(const modint rhs) const noexcept {
		return modint(*this) *= rhs;
	}
	constexpr modint operator/(const modint rhs) const noexcept {
		return modint(*this) /= rhs;
	}
	constexpr modint& operator+=(const modint rhs) noexcept {
		a += rhs.a;
		if (a >= Modulus) {
			a -= Modulus;
		}
		return *this;
	}
	constexpr modint& operator-=(const modint rhs) noexcept {
		if (a < rhs.a) {
			a += Modulus;
		}
		a -= rhs.a;
		return *this;
	}
	constexpr modint& operator*=(const modint rhs) noexcept {
		a = a * rhs.a % Modulus;
		return *this;
	}
	constexpr modint& operator/=(modint rhs) noexcept {
		u64 exp = Modulus - 2;
		while (exp) {
			if (exp % 2) {
				*this *= rhs;
			}
			rhs *= rhs;
			exp /= 2;
		}
		return *this;
	}
};
#pragma endregion

int main() {
	cin >> A >> B >> C >> D >> E;
	if(D > E){
		cout << min(A*D+B*E, min(2*C*E+A*(D-E), 2*C*max(D, E)));
	}else{
		cout << min(A*D+B*E, min(2*C*D+B*(E-D), 2*C*max(D, E)));
	}
}