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.Scanner;

public class Solution 
{
    public static void main(String args[])
    {
    int lengthOfArray,p,q,r;
    Scanner sc= new Scanner(System.in);
    
    //System.out.println("Enter the constants lengthOfArray,p,q,r: ");
    lengthOfArray = sc.nextInt();
    p=sc.nextInt();
    q=sc.nextInt();
    r=sc.nextInt();
    
   
    
    //System.out.println("Enter the number of elements in the  array :");
    //lengthOfArray =sc.nextInt();
    
    //System.out.println("Enter the elements of the array :");
    long[] array =new long[lengthOfArray];  
    for(int i=0;i<lengthOfArray;i++)
    {
        array[i]=sc.nextLong();
       
    }
    
    // calculating preffix max 
    long[] PMAX=new long[lengthOfArray];
    PMAX[0]=p*array[0];
    
    for(int i=1;i<lengthOfArray;i++)
    {
        PMAX[i]=Math.max(PMAX[i-1],array[i]*p);
        //System.out.println( PMAX[i] );
    }
    
    //calculating suffix max
    long[] SMAX = new long[lengthOfArray];
    SMAX[lengthOfArray-1]=r*array[lengthOfArray-1];
    
    for(int i=lengthOfArray-2;i>=0;i--)
    {
        SMAX[i]=Math.max(SMAX[i+1],array[i]*r);
        //System.out.println(SMAX[i] );
        
    }
    
    // Solving Marvalo Grunt Problem 
	 long ans=Long.MIN_VALUE;
	
	for(int i=0;i<lengthOfArray;i++)
	{
	ans = Math.max(ans,(PMAX[i]+(q*array[i])+SMAX[i]));
	//System.out.println( "answer" );
	
	//System.out.println( ans );
	}
	
	System.out.println( ans );
    
    }
    
    
    
}