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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Arrays;
import java.util.Scanner;

class Main{


	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);			//文字の入力
		int a = sc.nextInt();
		int b = sc.nextInt();
		int c = sc.nextInt();
		int x = sc.nextInt();
		int y = sc.nextInt();
		long ans = 0;

		int o = a*x + b*y;			//各ピザ

		int mi = Math.min(x, y);
		int ma = Math.max(x, y);
		int oo = 0;
		if(mi == x){
			oo = 2*c*mi +b*(ma-mi); 
		}else{
			oo = 2*c*mi +a*(ma-mi); 
		}
		int ooo = 2*c*ma;
		int[] data = {o,oo,ooo};
		Arrays.sort(data);
		
		System.out.println(data[0]);


	}





}

class UnionFind{
	int[] parent;
	public UnionFind(int N){
		this.parent = new int[N];
		for(int i = 0; i < N;i++){
			parent[i] = i;
		}
	}
	public int root(int x){			//根要素を取得
		if(parent[x] == x){			//初期状態
			return x;
		}else{
			return parent[x] = root(parent[x]);    	//再帰によって根を取得
		}
	}
	public void unite(int x,int y){ 	//統合メソッド
		if(same(x,y)){					//同じ根なら終了
			return;
		}
		int xroot = root(x);
		int yroot = root(y);
		parent[xroot] = yroot;			//統合
		return;

	}

	public boolean same(int x,int y){			//根が同じか（同じグループに属しているか）を判定
		return(root(x) == root(y));
	}
}



class Pair implements Comparable{
	int from;		//p
	int end;		//y
	int num;
	int bango;
	@Override
	public int compareTo(Object other) {
		Pair otherpair = (Pair)other;

		return end - otherpair.end;
	}
}