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 {
        Output out = new Output();
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;

        st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken()),
            p = Integer.parseInt(st.nextToken()),
            q = Integer.parseInt(st.nextToken()),
            r = Integer.parseInt(st.nextToken());
        
        st = new StringTokenizer(br.readLine());
        int[] a = new int[n];
        long[] max_on_right = new long[n];
        for (int i=0 ; i<n ; i++) {
            a[i] = Integer.parseInt(st.nextToken());
            max_on_right[i] = a[i];
        }
        
        
        max_on_right[n-1] = (long)r * max_on_right[n-1];
        for (int i=n-2 ; i>=0 ; i--) max_on_right[i] = Math.max(max_on_right[i]*r, max_on_right[i+1]);
        
        long sum = Long.MIN_VALUE;
        long left = (long)a[0]*p;
        for (int i=0 ; i<n ; i++) {
            left = Math.max(left, (long)a[i]*p);
            long current = (long)left + (long)a[i]*q + (long)max_on_right[i];
            sum = Math.max(current, sum);
            
        }
        
        out.println(sum);
        out.flush();
    }
    
    
    
}


class Output {
    private BufferedWriter out;

    Output() {
        this.out = new BufferedWriter(new OutputStreamWriter(System.out));
    }

    void print(String s) throws IOException {this.out.write(s);}
    void println(String s) throws IOException {this.out.write(s + "\n");}
    void println(long n) throws IOException {this.out.write(n + "\n");}
    void println(int n) throws IOException {this.out.write(n + "\n");}
    void flush () throws IOException {this.out.flush();}
}