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.io.*;
import java.util.*;

public class Main {
    public static void main(String[] args) throws java.lang.Exception {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        String[] l=br.readLine().split(" ");
        int n=Integer.parseInt(l[0]);
        long p=Long.parseLong(l[1]);
        long q=Long.parseLong(l[2]);
        long r=Long.parseLong(l[3]);
        String[] line=br.readLine().split(" ");
        long[] arr=new long[n];
        long[] suffixMax=new long[n];
        for(int i=n-1;i>=0;i--){
            arr[i]=Long.parseLong(line[i]);
            if(i<n-1){
                suffixMax[i]=Math.max(r*arr[i],suffixMax[i+1]);
            } else{
                suffixMax[i]=r*arr[i];
            }
        }
        long maxSum=Long.MIN_VALUE;
        long[] prefixMax = new long[n];
        prefixMax[0]=p*arr[0];
        for(int i=1;i<n;i++) prefixMax[i]=Math.max(prefixMax[i-1],p*arr[i]);
        for(int i=0;i<n;i++){
            maxSum=Math.max(maxSum,prefixMax[i]+(q*arr[i])+suffixMax[i]);
        }
        System.out.println(maxSum);
        
    }
}