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.Scanner;
import java.util.StringTokenizer;

public class milkOrders {
    public static void main(String[] args) throws IOException{

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        long p = Integer.parseInt(st.nextToken());
        long q = Integer.parseInt(st.nextToken());
        long r = Integer.parseInt(st.nextToken());

        long[] a = new long[n+1];

    st = new StringTokenizer(br.readLine());


        for(int i = 1; i <= n; i++){
            a[i] = Integer.parseInt(st.nextToken());
        }

        long[][] dp = new long[n+1][4];
        dp[1][0] = 0;
        dp[1][1] = p * a[1];
        dp[1][2] = p * a[1] + q * a[1];
        dp[1][3] = p * a[1] + q * a[1] + r * a[1];

        for(int i = 2; i <= n; i++){
            for(int j = 1; j <= 3; j++){
                if(j == 1){
                    dp[i][j] = Math.max(dp[i-1][1], p * a[i]);
                }
                else if(j == 2){
                    dp[i][j] = Math.max(Math.max(dp[i-1][2], dp[i-1][1] + q * a[i]), p * a[i] + q * a[i]);
                }
                else if(j == 3){
                    dp[i][j] = Math.max(Math.max(dp[i-1][3], dp[i-1][2] + r * a[i]), Math.max(dp[i-1][1] + q * a[i] + r * a[i], p * a[i] + q * a[i] + r * a[i]));
                }
            }
        }

        
	long max = 0 - Long.MAX_VALUE;


        for(int i = 1; i < dp.length; i++){
            max = Math.max(max, dp[i][3]);
        }
        out.println(max);
        out.close();

    }
}
