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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(
                new InputStreamReader(System.in));

        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        long p = Long.parseLong(st.nextToken());
        long q = Long.parseLong(st.nextToken());
        long r = Long.parseLong(st.nextToken());

        st = new StringTokenizer(br.readLine());
        long arr[]=new long[n];
        for (int i=0;i<n;i++){
            arr[i]=Long.parseLong(st.nextToken());
        }

        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]);
        }
//        System.out.println(Arrays.toString(pmax));
        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]);
        }
//        System.out.println(Arrays.toString(smax));
        long max=Long.MIN_VALUE;
        for (int i=0;i<n;i++){
            max=Math.max(max,pmax[i]+q*arr[i]+smax[i]);
        }
        System.out.println(max);

    }


}
