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.
//I AM THE CREED
/* //I AM THE CREED
/* package codechef; // don't place package name! */
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.StringTokenizer; 
import java.util.*;
import java.awt.Point;
public class Main{
    public static void main(String[] args) throws IOException 
    { 
        Scanner input = new Scanner(System.in);
        while(input.hasNext()){
            int n=input.nextInt();
            long p=input.nextLong();
            long q=input.nextLong();
            long r=input.nextLong();
            long[] l=new long[n];
            for(int i=0;i<n;i++){
                l[i]=input.nextLong();
            }
            long[][] suffix=new long[n][2];
            long[][] prefix=new long[n][2];
            suffix[n-1][0]=l[n-1];
            suffix[n-1][1]=l[n-1];
            prefix[0][0]=l[0];
            prefix[0][1]=l[0];
            for(int i=n-2;i>=0;i--){
                suffix[i][0]=Math.min(l[i], suffix[i+1][0]);
                suffix[i][1]=Math.max(l[i], suffix[i+1][1]);
            }
            for(int i=1;i<n;i++){
                prefix[i][0]=Math.min(l[i], prefix[i-1][0]);
                prefix[i][1]=Math.max(l[i], prefix[i-1][1]);
            }
            long ans=Long.MIN_VALUE;
            for(int i=0;i<n;i++){
                for(int j=0;j<=1;j++){
                    for(int k=0;k<=1;k++){
                        ans=Math.max(ans, p*prefix[i][j]+l[i]*q+r*suffix[i][k]);
                    }
                }
            }
            System.out.println(ans);
        }
    }
    
}