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 Main {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt(),p= scan.nextInt(),q=scan.nextInt(),r=scan.nextInt();
        long arr[]= new long[n];
        for(int i=0;i<n;i++) {
            arr[i] = scan.nextInt();
        }
        long pmax[] = new long[n];
        long smax[] = new long[n];
        long temp= Long.MIN_VALUE;
        for(int i=0;i<n;i++) {
            temp=Math.max(temp,p*arr[i]);
            pmax[i]=temp;
        }
        long temp2= Long.MIN_VALUE;
        for(int i=n-1;i>=0;i--) {
            temp2=Math.max(temp2,r*arr[i]);
            smax[i]=temp2;
        }
        long ans= Long.MIN_VALUE;
        long left,right;
        for(int j=0;j<n;j++) {
            left=pmax[j];
            right=smax[j];
            ans=Math.max(ans,(left+q*arr[j]+right));
        }
        System.out.print(ans);
    }
}


