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.Scanner;
public class MarvoloGauntsRing {

	public static void main(String[] args) {
		Scanner s=new Scanner(System.in);
		int n=s.nextInt();
		long p=s.nextLong();
		long q=s.nextLong();
		long r=s.nextLong();
		long arr[]=new long[n];
		for(int i=0;i<arr.length;i++) {
			arr[i]=s.nextLong();
		}
      long prefix[]=new long[n];
      long maxleft=Long.MIN_VALUE;
      for(int i=0;i<n;i++) {
    	  if(p*arr[i]>maxleft) {
    		  prefix[i]=p*arr[i];
    		  maxleft=p*arr[i];
    	  }else {
    		  prefix[i]=maxleft;
    	  }
      }
      
      long suffix[]=new long[n];
      long maxright=Long.MIN_VALUE;
      for(int i=n-1;i>=0;i--) {
    	  if(r*arr[i]>maxright) {
    		  suffix[i]=r*arr[i];
    		  maxright=r*arr[i];
    	  }else {
    		  suffix[i]=maxright;
    	  }
      }
       long sum=Long.MIN_VALUE;
		for(int j=0;j<arr.length;j++) {
			maxleft=prefix[j];
			maxright=suffix[j];
		 sum=Math.max(maxleft+q*arr[j]+maxright,sum);
		 
		}
        System.out.println(sum);
	}

}
