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 Java7 (OpenJDK 1.7.0) (judge verdict: Accepted)

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.lang.reflect.Array;
import java.math.BigDecimal;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.StringTokenizer;

public class Main {
	public static void main(String[] args) {
		try {
			
			BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

			String[] hyozyun = br.readLine().split(" ");
			int A =Integer.parseInt(hyozyun[0]);
			int B =Integer.parseInt(hyozyun[1]);
			int C =Integer.parseInt(hyozyun[2]);
			int X =Integer.parseInt(hyozyun[3]);
			int Y =Integer.parseInt(hyozyun[4]);
			
			// 合計金額を設定
			int sum = 0;
			// Cを２枚買ったほうが安い場合
			if (A + B < 2 * C) {
				sum = A * X + B * Y;
			} else {
			// それぞれ買ったほうが安い場合
				// 枚数少ないほうをすべて買って、プラス足りない分購入
				if (X > Y) {
					sum = C * 2 * Y + (X - Y) * A;
				} else {
					sum = C * 2 * X + (Y - X) * B;
				}
			}
			// XとYの枚数が多いほうの分だけ、Cを買った場合の値段
			int compare = 0;
			if (X > Y) {
				compare = X * C * 2;
			} else {
				compare = Y * C * 2;
			}
			// 上記の計算結果のほうが安い場合はそちらを採用
			if (sum > compare) {
				sum = compare;
			}
			
			System.out.println(sum);
		} catch (Exception e) {
			e.printStackTrace();
		}
	}
}