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.*;

public class Main
{
	public static void main (String[] args ) throws java.lang.Exception
	{
	
   Scanner sc=new Scanner(System.in);
   int n=sc.nextInt();
   long p=sc.nextLong();
   long q=sc.nextLong();
   long r=sc.nextLong();
   long a[]=new long[(int)n];

   for(int i=0;i<n;i++){
   	 a[i]=sc.nextInt();
   }
   
        long pmax[]=new long[(int)n];
         pmax[0]=a[0]*p;
        
        for(int i=1;i<=n-1;i++){
            pmax[i]=Math.max(pmax[i-1],a[i]*p);
        }
        
         long Smax[]=new long[(int)a.length];
         Smax[n-1]=a[n-1]*r;
        
        for(int i=n-2;i>=0;i--){
            Smax[i]=Math.max(Smax[i+1],a[i]*r);
        }
        
    long ans= Long.MIN_VALUE;
    for(int i=0;i<=n-1;i++){
        ans=Math.max(ans,pmax[i]+q*a[i]+Smax[i]);
    }
    
    System.out.println(ans);
       
	}
}
