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.
// package Greedy;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class HarryPotterPotion {
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        String line[]=br.readLine().split(" ");
        int n=Integer.parseInt(line[0]);
        int p=Integer.parseInt(line[1]);
        int q=Integer.parseInt(line[2]);
        int r=Integer.parseInt(line[3]);
        line=br.readLine().split(" ");
        long a[][]=new long[4][n];
        for(int i=0;i<n;i++){
            a[0][i]=Long.parseLong(line[i]);
            a[1][i]=a[0][i];
            a[2][i]=a[0][i];
            a[3][i]=a[0][i];
        }
        long dp[][]=new long[n+1][3];
        dp[0][0]=dp[0][1]=dp[0][2]=Long.MIN_VALUE;
        for(int i=1;i<=n;i++){
            long curr_p=a[1][i-1];
            long curr_q=a[2][i-1];
            long curr_r=a[3][i-1];
            dp[i][0]=Math.max(dp[i-1][0],curr_p*p);
            dp[i][1]=Math.max(dp[i-1][1],dp[i][0]+curr_q*q);
            dp[i][2]=Math.max(dp[i-1][2],dp[i][1]+curr_r*r);
//            System.out.println(dp[i][2]);
        }
        System.out.println(dp[n][2]);
    }
}
