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.ArrayList;
import java.util.Scanner;

public class CodeForces {
	
	
	
   
    public static void main(String args[])
    {
    	Scanner scan = new Scanner(System.in);
    	int n = scan.nextInt();
    	long p = scan.nextLong();
    	long q = scan.nextLong();
    	long r = scan.nextLong();
    	int[] a = new int[n+1];
    	for(int i=1;i<=n;i++)
    	{
    		a[i] = scan.nextInt();
    	}
    	
    	int[] left_min = new int[n+1];
    	int[] left_max = new int[n+1];
    	
    	left_min[1] = left_max[1] = a[1];
    	
    	for(int i=2;i<=n;i++)
    	{
    		left_min[i] = Math.min(left_min[i-1], a[i]);
    		left_max[i] = Math.max(left_max[i-1], a[i]);
    	}
    	
    	int[] right_min = new int[n+1];
    	int[] right_max = new int[n+1];
    	
    	right_min[n] = right_max[n]= a[n];
    	
    	for(int i = n-1 ; i>=1 ; i--)
    	{
    		right_max[i] = Math.max(right_max[i+1], a[i]);
    		right_min[i] = Math.min(right_min[i+1], a[i]);
    		
    	}
    	
    	long ans = p*a[1]+q*a[1]+r*a[1];
    	
    	for(int i=1;i<=n;i++)
    	{
    		long y = q*a[i];
    		long x = (p > 0 ? p*left_max[i] : p*left_min[i]);
    		long z = (r > 0 ? r*right_max[i] : r*right_min[i]);
    		
    		ans = Math.max(ans,  (x+y+z));
    	}
    	System.out.println(ans);
    }
}
