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 whatever; // don't place package name! */

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

/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
		Scanner sc = new Scanner(System.in);
		int n;
		long p, q, r;
		n = sc.nextInt();
		p = sc.nextLong();
		q = sc.nextLong();
		r = sc.nextLong();
		
		long arr[] = new long[n];
		for(int ind = 0; ind < n; ind++) {
			arr[ind] = sc.nextLong();
		}
		
		long pmax[] = new long[n];
		long smax[] = new long[n];
		
		pmax[0] = p * arr[0];
		
		for(int ind = 1; ind < n; ind++) {
			pmax[ind] = Math.max(pmax[ind - 1], p * arr[ind]);
		}
		
		smax[n - 1] = r * arr[n - 1];
		for (int ind = n - 2; ind >= 0; ind--) {
			smax[ind] = Math.max(smax[ind + 1], r * arr[ind]);
		}
		
		long ans = Long.MIN_VALUE;
		for (int ind = 0; ind < n; ind++) {
			ans = Math.max(ans, pmax[ind] + (q * arr[ind]) + smax[ind]);
		}
		System.out.println(ans);
	}
}