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  MarvoloGauntsRing
{
	public static void main (String[] args) throws java.lang.Exception
	{
		Scanner scannerInput = new Scanner(System.in);
		int n = scannerInput.nextInt();
		long p = scannerInput.nextLong();
		long q = scannerInput.nextLong();
		long r = scannerInput.nextLong();
		long[] arr = new long[n];
		long[] prefixMax = new long[n];
		long[] suffixMax = new long[n];
		for(int i =  0 ; i < n ; i++){
			arr[i] = scannerInput.nextLong();
		}
		prefixMax[0] = p*arr[0];
		suffixMax[n-1] = r*arr[n-1];
		for(int i = 1 ;  i < n ; i++){
			prefixMax[i] = Long.max(arr[i]*p , prefixMax[i - 1]);
		}
		for(int i = n - 2 ;  i >= 0 ; i--){
			suffixMax[i] = Long.max(arr[i]*r , suffixMax[i + 1]);
		}
		
		
		long ans = Long.MIN_VALUE;
		for(int i = 0 ; i < n ;  i++){
			ans = Long.max(ans , prefixMax[i] + (q*arr[i]) + suffixMax[i]);
		}
		System.out.println(ans);
	}
}