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 //do not write package name here */
/*package whatever //do not write package name here */
// marvolo gaunt ring

import java.io.*;
import java.util.Scanner;

public class File {
	public static void main (String[] args) 
	{
	Scanner sc=new Scanner(System.in);
	long p,q,r;
	int n;
    //reading the number of elements from the that we want to enter  
    n=sc.nextInt(); 
	p=sc.nextLong();
	q=sc.nextLong();
	r=sc.nextLong();
	long arr[]=new long[n];
	for(int i=0; i<n; i++)
	{
	   arr[i]=sc.nextLong(); 
	}
	
	long parr[]=new long[n];
	long curmaxp;
	curmaxp=((long)(p*arr[0]));
	for(int i=1; i<n; i++)
	{
	    curmaxp=Math.max((long)(p*arr[i]),curmaxp);
	    parr[i]=curmaxp;
	}
	parr[0]=((long)(p*arr[0]));
	
	long sarr[]=new long[n];
	long curmax;
	curmax=((long)(r*arr[n-1]));
	for(int i=n-2; i>=0; i--)
	{
	    curmax=Math.max((long)(r*arr[i]),curmax);
	    sarr[i]=curmax;
	}
	sarr[n-1]=((long)(r*arr[n-1]));
	
	long curx;
	long maxSoFar= Long.MIN_VALUE;
	long x;
	for(int j=0; j<n; j++)
	{
	    curx=(parr[j]+(long)(q*arr[j])+sarr[j]);
	    maxSoFar=Math.max(maxSoFar,curx);
	}
	x=maxSoFar;
    System.out.println(x);
	}
}