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 codechef; // 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 Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		try {
		    Scanner sc=new Scanner(System.in);
		    int n=sc.nextInt();
		    long p=sc.nextLong();
		    long q=sc.nextLong();
		    long r=sc.nextLong();
		    
		    long arr[]=new long[n];
		    for(int i=0;i<n;i++){
		        arr[i]=sc.nextLong();
		    }
		    long prefix_max[]=new long[n];
		    long suffix_max[]=new long[n];
		    
		    prefix_max[0]=p*arr[0];
		    for(int i=1;i<n;i++){
		        prefix_max[i]=Math.max(prefix_max[i-1],p*arr[i]);
		    }
		    
		    suffix_max[n-1]=r*arr[n-1];
		    for(int i=n-2;i>=0;i--){
		        suffix_max[i]=Math.max(suffix_max[i+1],r*arr[i]);
		    }
		    long ans=Long.MIN_VALUE;
		    for(int j=0;j<n;j++){
		        ans=Math.max(ans,prefix_max[j]+ (arr[j]*q) + suffix_max[j]);
		    }
		    System.out.println(ans);
		} catch(Exception e) {
		    System.out.println(e);
		}
	}
}
