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.Scanner;

public class _A1_MaxSum
{
	public static void main(String[] args)
	{
		Scanner input = new Scanner(System.in);
		Scanner sc = new Scanner(System.in);
		int arrSize = sc.nextInt();
		long p = sc.nextLong();
		long q = sc.nextLong();
		long r = sc.nextLong();
		Long[] arr = new Long[arrSize];

		for (int i = 0; i < arrSize; i++)
			arr[i] = sc.nextLong();

		long[] preMax = new long[arrSize];
		long[] sufMax = new long[arrSize];
		long maxRes = Long.MIN_VALUE;

		for (int i = 0; i < arrSize; i++)
		{
			long temp = arr[i] * p;
			maxRes = Math.max(temp, maxRes);
			preMax[i] = maxRes;

		}
		maxRes = Long.MIN_VALUE;

		for (int i = arrSize - 1; i >= 0; i--)
		{
			long temp = arr[i] * r;
			maxRes = Math.max(temp, maxRes);
			sufMax[i] = maxRes;
		}

		long result = Long.MIN_VALUE;
		for (int i = 0; i < arrSize; i++)
		{
			long lt = preMax[i];
			long rt = sufMax[i];

			if (result < rt + lt + arr[i] * q)
			{
				result = rt + lt + arr[i] * q;
			}
		}
		System.out.println(result);
	}

}