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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#define MIN_P 1
#define MAX_P 5000
#define MIN_X 1
#define MAX_X 100000

int main (void) {
	int a, b, c, x, y, min_price, nx, ny;
	scanf("%d %d %d %d %d[^\n]", &a, &b, &c, &x, &y);
	if (a < MIN_P || a > MAX_P || b < MIN_P || b > MAX_P || c < MIN_P || c > MAX_P || x < MIN_X || x > MAX_X || y < MIN_X || y > MAX_X) return -1;
	if (x >= y) {
		min_price = (a + b <= 2 * c) ? (a + b) * y : 2 * c * y;
		nx = x - y;
		min_price = a * nx + min_price;
		min_price = (min_price <= 2 * x * c) ? min_price : 2 * x * c;
	} else {
		min_price = (a + b <= 2 * c) ? (a + b) * x : 2 * c * x;
		ny = y - x;
		min_price = b * ny + min_price;
		min_price = (min_price <= 2 * y * c) ? min_price : 2 * y * c;
	}
	printf("%d\n", min_price);
	return 0;
}