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.*;
import java.lang.*;
import java.io.*;
 
 public class Ideone
{
	public static void main (String[] args) throws java.lang.Exception
	{
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		Long arr[]=new Long[n];
 
		int p=sc.nextInt();
		int q=sc.nextInt();
		int r=sc.nextInt();
 
		sc.nextLine();
		for(int t=0;t<n;t++)
		arr[t]=sc.nextLong();
 
 
 
		Long preMax[]=new Long[n];
		Long sufMax[]=new Long[n];
		Long ans=Long.MIN_VALUE;
 
		preMax[0]=p*arr[0];
		for(int i=1;i<n;i++){
		preMax[i]=Math.max(preMax[i-1],p*arr[i]);
		//System.out.println(preMax[i]);
		}
 
		sufMax[n-1]=r*arr[n-1];
		for(int k=n-2;k>=0;k--){
		sufMax[k]=Math.max(sufMax[k+1],r*arr[k]);
		//System.out.println(sufMax[k]);
		}
		for(int j=0;j<n;j++){
		Long x=preMax[j]+sufMax[j]+q*arr[j];
		//System.out.println(x);
		ans=Math.max(ans,x);
 
		}
 
		System.out.println(ans);
	}
 
 
 
 
}