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.
/*
ID: billy_b2
LANG: JAVA
TASK: ring
*/
import java.io.*;
import java.util.*;

public class ring
{
	public static long max(long a, long b) {return a > b ? a : b;}
	
	public static void main(String[] args) throws IOException
	{
		Scanner in = new Scanner(System.in);
		PrintStream out = System.out;
		
		int n = in.nextInt();
		long p = in.nextLong();
		long q = in.nextLong();
		long r = in.nextLong();
		long[] arr = new long[n];
		for(int i = 0; i < n; i++) arr[i] = in.nextLong();
		
		long[][] dp = new long[n][3]; //dp[i][j] = using only first i entries of arr, only using the first j variables in (p, q, r)
		dp[0][0] = p * arr[0];
		dp[0][1] = q * arr[0] + dp[0][0];
		dp[0][2] = r * arr[0] + dp[0][1];
		for(int i = 1; i < n; i++) dp[i][0] = max(dp[i-1][0], arr[i] * p); //fill in dp for the variable p
		for(int i = 1; i < n; i++) dp[i][1] = max(dp[i-1][1], dp[i][0] + arr[i] * q); //fill in dp for the variables p+q
		for(int i = 1; i < n; i++) dp[i][2] = max(dp[i-1][2], dp[i][1] + arr[i] * r); //fill in dp for the variables p+q+r
		
		/*
		for(int i = 0; i < 3; i++)
		{
			for(int j = 0; j < n; j++)
				out.print(dp[j][i] + " ");
			out.println();
		}
		*/
		
		out.println(dp[n-1][2]);
		out.close();
	}
}
