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.*;
import java.io.*;

public class Marvolo_Gaunts_Ring
{
	public static void main(String[] args) throws IOException 
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw=new PrintWriter(System.out);

		int t=1;

		while(t-->0)
		{

			String str[]=br.readLine().trim().split(" ");
			int n=Integer.parseInt(str[0]);
			long p=Long.parseLong(str[1]);
			long q=Long.parseLong(str[2]);
			long r=Long.parseLong(str[3]);

			str=br.readLine().trim().split(" ");
			
			long b[]=new long[n];
			long a[]=new long[n];
			long c[]=new long[n];
			for(int i=0;i<n;i++)
			{
				b[i]=Long.parseLong(str[i]);
			}

			a[0]=p*b[0];
			for(int i=1;i<n;i++)
			{
				a[i]=Math.max(a[i-1],b[i]*p);
			}

			c[n-1]=r*b[n-1];
			for(int i=n-2;i>=0;i--)
			{
				c[i]=Math.max(c[i+1],b[i]*r);
			}

			long ans=Long.MIN_VALUE;

			for(int i=0;i<n;i++)
			{
				long x=q*b[i]+a[i]+c[i];

				ans=Math.max(ans,x);
			}

			pw.println(ans);
		}
		pw.flush();
	}
}