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 Codeforces855B {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long p,q,r;
        p=sc.nextLong();
        q=sc.nextLong();
        r=sc.nextLong();
        long[] array = new long[n];
        for (int i =0; i<n;i++){
            array[i]=sc.nextLong();
        }
        long sum=Long.MIN_VALUE;

        //Prefix Max
        long[] prefix_max = new long[array.length];
        for (int i=0; i<array.length;i++){
            if (i==0){
                prefix_max[i]= p*array[i];
            }
            else {
                prefix_max[i] = Math.max(prefix_max[i - 1], p*array[i]);
            }
        }

        //Suffix Max
        long[] suffix_max = new long[array.length];
        for (int i= array.length-1; i>=0;i--){
            if(i== array.length-1){
                suffix_max[i] = r*array[i];
            }
            else {
                suffix_max[i] = Math.max(suffix_max[i+1],r*array[i]);
            }
        }

        //for middle term, finding max sum
        for(int mid = 0; mid <= array.length-1;mid++){
            sum =Math.max(sum,prefix_max[mid]+q*array[mid]+suffix_max[mid]);
        }
        System.out.println(sum);
    }
}
