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.
/* package whatever; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Ideone
{
	public static void main (String[] args) throws java.lang.Exception
	{
	Scanner sc=new Scanner(System.in);
        int n;
        long p,q,r;
        n=sc.nextInt();
        p=sc.nextLong();
        q=sc.nextLong();
        r=sc.nextLong();
        long[] a=new long[n];
        for(int i=0;i<n;i++)a[i]=sc.nextLong();

        long[] arr1=new long[n];
        long[] arr2=new long[n];
        long[] arr3=new long[n];

        arr1[0]=a[0]*p;
        for(int i=1;i<n;i++){
            arr1[i]=Long.max(arr1[i-1],a[i]*p);
        }

        arr2[0]=arr1[0]+a[0]*q;
        for(int i=1;i<n;i++){
            arr2[i]=Long.max(arr2[i-1],a[i]*q+arr1[i]);
        }

        arr3[0]=arr2[0]+a[0]*r;
        for(int i=1;i<n;i++){
            arr3[i]=Long.max(arr3[i-1],a[i]*r+arr2[i]);
        }
        System.out.println(arr3[n-1]);

	}
}