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;
//https://github.com/sourabhkanojia/CodeForces/blob/master/855B%20-%20Marvolo%20Gaunt's%20Ring.java
public class ThreeSumsAgain
{
	public static void main(String[] args)
	{
		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[] a = new long[n];
		
		for (int i = 0; i < n; i++)
		{
			a[i] = sc.nextLong();
		}

		long[] dp1 = new long[n];
		long[] dp2 = new long[n];
		long[] dp3 = new long[n];

		dp1[0] = a[0] * p;
		for (int i = 1; i < n; i++)
		{
			dp1[i] = Long.max(dp1[i - 1], a[i] * p);
		}

		dp2[0] = dp1[0] + a[0] * q;
		for (int i = 1; i < n; i++)
		{
			dp2[i] = Long.max(dp2[i - 1], a[i] * q + dp1[i]);
		}

		dp3[0] = dp2[0] + a[0] * r;
		for (int i = 1; i < n; i++)
		{
			dp3[i] = Long.max(dp3[i - 1], a[i] * r + dp2[i]);
		}
		System.out.println(dp3[n - 1]);
	}

}
