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.*;
import java.lang.*;
import java.math.*;
public class MarvoloRing{

     public static void main(String []args){
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        long[] arr = new long[n];
        long x = s.nextLong();
        long y = s.nextLong();
        long z = s.nextLong();
        
        for(int i=0;i<n;i++)
            arr[i]=s.nextLong();
        
        
        
        long[] pmax = new long[n];
        pmax[0] = x*(arr[0]);
        
        for(int i=1;i<n;i++)
            pmax[i] = Long.max(pmax[i-1],x*arr[i]);
            
            
        long[] smax = new long[n];
        smax[n-1] = z*arr[n-1];
        for(int i = n-2;i>=0;i--)
            smax[i] = Long.max(smax[i+1],z*arr[i]);
            
        Long maxSum = Long.MIN_VALUE;
        
        /*
            If i<j<k, then
            i runs from 1 to n-2, because 1st element cannot be middle element.
            1st element's index < middle element's index < last element's index.
            
            for(int i = 1;i<= n-2;i++)
            {
                maxSum = Math.max(maxSum,pmax[i-1]+(y*arr[i])+smax[i+1]);
            }
            pmax[i-1]-->corresponds to constraint i and 
            smax[i+1]-->corresponds to constraint k
            because i < j < k
            
        */
        
        // This is when i <= j <= k
        for(int i = 0;i<= n-1;i++)
        {
            maxSum = Long.max( maxSum, pmax[i]+(y*arr[i])+smax[i] );
        }
        System.out.println(maxSum);
     }
}