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.util.*;

//Maximize the expression p*arr[i] + q*arr[j] + r*arr[k]

public class MaximizeExpression{
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n= in.nextInt();
        int p= in.nextInt();
        int q= in.nextInt();
        int r= in.nextInt();
        long []arr= new long[n];
        for(int i=0; i<n; i++){
            arr[i]= in.nextLong();
        }
        long ans= maximize(p,q,r,arr);
        System.out.println(ans);
        in.close();
    }

    private static long maximize(int p, int q, int r, long[] arr) {
        int n= arr.length;
        long []prefix= new long[n];
        long []suffix= new long[n];
        long ans=Long.MIN_VALUE;
        //calculate prefix max
        prefix[0]=arr[0]*p;
        for(int i=1; i<n; i++){
            prefix[i]= Math.max(prefix[i-1], p*arr[i]);
           
        }
        //calculate suffix max
        suffix[n-1]= arr[n-1]*r;
        for(int i=n-2; i>=0; i--){
            suffix[i]= Math.max(suffix[i+1], r*arr[i]);
            
        }
        //Calculate ans
        for(int i=0; i<n; i++){
            ans=Math.max(ans,prefix[i]+q*arr[i]+suffix[i]);
        }
        return ans;
    }
}