Professor Dumbledore is helping Harry destroy the Horcruxes. He went to Gaunt Shack as he suspected a Horcrux to be present there. He saw Marvolo Gaunt's Ring and identified it as a Horcrux. Although he destroyed it, he is still affected by its curse. Professor Snape is helping Dumbledore remove the curse. For this, he wants to give Dumbledore exactly x drops of the potion he made. 

Value of x is calculated as maximum of p·ai + q·aj + r·ak for given p, q, r and array a1, a2, ... an such that 1 ≤ i ≤ j ≤ k ≤ n. Help Snape find the value of x. Do note that the value of x may be negative.

Input

First line of input contains 4 integers n, p, q, r ( - 109 ≤ p, q, r ≤ 109, 1 ≤ n ≤ 105).

Next line of input contains n space separated integers a1, a2, ... an ( - 109 ≤ ai ≤ 109).

Output

Output a single integer the maximum value of p·ai + q·aj + r·ak that can be obtained provided 1 ≤ i ≤ j ≤ k ≤ n.

Examples

Input

5 1 2 3
1 2 3 4 5


Output

30


Input

5 1 2 -3
-1 -2 -3 -4 -5


Output

12

Note

In the first sample case, we can take i = j = k = 5, thus making the answer as 1·5 + 2·5 + 3·5 = 30.

In second sample case, selecting i = j = 1 and k = 5 gives the answer 12.
 import java.util.Scanner;
 
public class Pa {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int a1 = sc.nextInt();
        long a2 = sc.nextLong();
        long a3 = sc.nextLong();
        long a4 = sc.nextLong();
 
        long[] array = new long[a1];
        for (int i = 0; i < a1; i++) {
            array[i] = sc.nextInt();
        }
        long[] maxi = new long[a1];
        maxi[0] = a2 * array[0];
 
        for(int i = 1 ; i<a1; i++){
            maxi[i] = Math.max(maxi[i-1], a2 * array[i]);
        }
 
        long[] maxi2 = new long[a1];
        maxi2[a1-1] = a4 * array[a1-1];
        for(int i = a1-2; i >= 0; i--){
            maxi2[i] = Math.max(maxi2[i+1],a4 * array[i] );
        }
        long ansi = Long.MIN_VALUE;
        for(int i = 0; i< a1; i++){
            ansi = Math.max(ansi , maxi[i]+(a3 * array[i])+maxi2[i]);
        }
 
        System.out.println(ansi);
 
 
    }
}