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.*;
import java.io.*;
import java.lang.Math;

public class solution{
	public static void main(String[] args) throws IOException{
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		long p = s.nextLong();
		long q = s.nextLong();
		long r = s.nextLong();
		long[] array = new long[n];
		for(int i=0;i<n;i++) array[i] = s.nextLong();
		long[] maxLeft = new long[n];
		maxLeft[0] = (long)p*array[0];
		for(int i=1;i<n;i++){
			maxLeft[i] = (long)Math.max(maxLeft[i-1],(long)p*array[i]);
		}
		long[] maxRight = new long[n];
		maxRight[n-1] = (long)r*array[n-1];
		for(int i=n-2;i>=0;i--){
			maxRight[i] = (long)Math.max(maxRight[i+1],(long)r*array[i]);
		}
		long result = Long.MIN_VALUE;
		for(int i=0;i<n;i++){
			result = (long)Math.max(result,(long)maxLeft[i]+maxRight[i]+q*array[i]);
		}
		System.out.println(result);
	}
}