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.*;

public class Marvolo_Gount {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc=new Scanner(System.in) ;
		int n=sc.nextInt();
		long p=sc.nextLong();
		long q=sc.nextLong();
		long r=sc.nextLong();
		long arr[]=new long[n];
		for(int i=0;i<n;i++)
		{
			arr[i]=sc.nextLong();
		}
		
		long prefix_min[]=new long[n];
		long prefix_max[]=new long[n];
		long suffix_max[]=new long[n];
		long suffix_min[]=new long[n];
		
		//prefix_min
		prefix_min[0]=arr[0];
		for(int i=1;i<n;i++)
		{
			prefix_min[i]=Math.min(prefix_min[i-1], arr[i]);
		}
		
		//prefix_max
		prefix_max[0]=arr[0];
		for(int i=1;i<n;i++)
		{
			prefix_max[i]=Math.max(prefix_max[i-1], arr[i]);
		}
		
		//suffix_min
		suffix_min[n-1]=arr[n-1];
		for(int i=n-2;i>=0;i--)
		{
			suffix_min[i]=Math.min(arr[i], suffix_min[i+1]);
		}
		
		//suffix_max
		suffix_max[n-1]=arr[n-1];
		for(int i=n-2;i>=0;i--)
		{
			suffix_max[i]=Math.max(arr[i], suffix_max[i+1]);
		}
		
		long max=(long)Math.pow(2,63),left=0,right=0;
		max*=(-1);
		for(int i=0;i<n;i++)
		{
			if(p<=0)
				left=(p*prefix_min[i]);
			else
				left=(p*prefix_max[i]);
			if(r<=0)
				right=(r*suffix_min[i]);
			else
				right=(r*suffix_max[i]);
			
			max=Math.max((left+right+(q*arr[i])), max);
		}
		
		System.out.println(max);
	}

}