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.io.*;
import java.util.*;
public class Contest
{
	public static void main(String args[])
	{
		Scanner scanner = new Scanner(System.in);
		int numberOfElements = scanner.nextInt();
		int p = scanner.nextInt();
		int q = scanner.nextInt();
		int r = scanner.nextInt();
		
		int arr[] = new int[numberOfElements+1];
		for (int i = 1; i <= numberOfElements; i++)
		{
			arr[i] = scanner.nextInt();
		}
		
		long cache[][] = new long[3][numberOfElements+1];
		
		cache[0][0] = cache[1][0] = cache[2][0] = Long.MIN_VALUE;
		for (int i = 1; i <= numberOfElements; i++) {
			cache[0][i] = max(cache[0][i-1],arr[i]*(long)p);
		}
		for (int i = 1; i <= numberOfElements; i++)
		{
			cache[1][i] = max(cache[1][i-1],arr[i]*(long)q  + cache[0][i]);
		}
		for (int i = 1; i <= numberOfElements; i++)
		{
			cache[2][i] = max(cache[2][i-1],arr[i]*(long)r + cache[1][i]);
		}
		System.out.println(cache[2][numberOfElements]);
	}
	public static long max(long a,long b)
	{
		return a>b?a:b;
	}
}