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.
/*
5 1 2 3
1 2 3 4 5

*/

import java.util.*;
import java.lang.Math.*;

public class ring_855
{
    public static double max(double a, double b)
    {
        double mx = (a > b) ? a : b;
        return(mx);
    }
    
    public static void main(String[] args)
    {
        String newline = System.getProperty("line.separator");
        Scanner ins = new Scanner(System.in);
        int  n = ins.nextInt();
        long p = ins.nextInt();
        long q = ins.nextInt();
        long r = ins.nextInt();
        
        long a = ins.nextInt();
        long best1 = p * a;
        long best2 = p*a + q*a;
        long best3 = p*a  + q*a + r*a;       
        
        
        for (int i=1; i<n;i++)
        {
            long tmp;
            a = ins.nextInt();
            tmp = p*a;
            if (tmp > best1)
            {
                best1 = tmp;
            }
            tmp = best1 + (q * a);
            if (tmp > best2)
            {
                best2 = tmp;
            }
            tmp = best2 + (r * a);
            if (tmp > best3)
            {
                best3 = tmp;
            }
            
            
            
        }
        
        System.out.print(best3);
        
        
        
    }
    
    
}
