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 MaxExpression {

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long p = sc.nextLong();
        long q = sc.nextLong();
        long r = sc.nextLong();

        long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = sc.nextInt();
        }

        System.out.println(func(arr,p,q,r));
    }

   static public long func(long arr[], long p, long q, long r) {
        int n = arr.length;

        long pmax[] = new long[n];
        pmax[0]=p*arr[0];
        for (int i = 1; i < n; i++)
        {
            pmax[i]=Math.max(pmax[i-1],p*arr[i]);
        }
        long smax[] = new long[n];
        smax[n-1]= r*arr[n-1];
        for(int i=n-2;i>=0;i--)
        {
          smax[i]=Math.max(smax[i+1],r*arr[i]);
        }

       long ans=Long.MIN_VALUE;
        for(int i=0;i<n;i++)
        {
            ans = Math.max(ans, (pmax[i]+q*arr[i]+smax[i]));
        }
        return ans;
    }
}
