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.
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import java .io.*;
import java .util.*;


public class Main
{
	public static void main(String[] args)throws Exception
	{
		Scanner s=new Scanner (System.in);
		long n =s.nextLong();
		long p=s.nextLong();
		long q=s.nextLong();
		long r=s.nextLong();
		long a[]=new long[(int)n];
		for(int i=0;i<n;i++)
		{
		    a[i]=s.nextLong();
		    
		}
		long sum=Long.MIN_VALUE;
		long pMax[]=new long[(int)n],sMax[]=new long[(int)n];
		pMax[0]=p*a[0];
		sMax[(int)n-1]=r*a[(int)n-1];
		
		for(int i=1;i<n;i++)
		{
		    pMax[i]=Math.max(pMax[i-1],p*a[i]);
		    //	System.out.println(psum[i]);
		}
		for(int i=(int)n-2;i>=0;i--)
		{
		    sMax[i]=Math.max(sMax[i+1],r*a[i]);
		    	//System.out.println(ssum[i]);
		}
		for(int i=0;i<n;i++)
		{
		    sum=Math.max(sum,(pMax[i]+q*a[i]+sMax[i]));
		    //	System.out.println(sum);
		}
		System.out.println(sum);
		
		
	}
}
