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.
//package codeForce4;

import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;
import java.util.TreeMap;

public class drop {
	
	public static void main(String args[])
	{
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		ArrayList<Long> constant = new ArrayList<Long>();
		long p  = in.nextLong();
		long q = in.nextLong();
		long r = in.nextLong();
		
		
		long[] values = new long[n];
		
		for(int i = 0; i < n; i++)
		{
			long value = in.nextLong();
			values[i] = value;
		}
		
		long[] P = new long[n];
		P[0] = p * values[0];
		long max = p*values[0];
		for(int i = 1; i < n; i++)
		{
			long current = p * values[i];
			if(current > max)
			{
				max = current;
				P[i] = max;
			}
			else
			{
				P[i] = max;
			}
		}
		
		long[] Q = new long[n];
		max = P[0] + q*values[0];
		Q[0] = max;
		for(int i = 1; i < n; i++)
		{
			long current = q * values[i] + P[i];
			if(current > max)
			{
				max = current;
				Q[i] = max;
			}
			else
			{
				Q[i] = max;
			}
		}
		
		long[] R = new long[n];
		max = Q[0] + r*values[0];
		R[0] = max;
		for(int i = 1; i < n; i++)
		{
			long current = r * values[i] + Q[i];
			if(current > max)
			{
				max = current;
				R[i] = max;
			}
			else
			{
				R[i] = max;
			}
		}
		
		System.out.println(R[n-1]);
		
	}

}