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.math.*;

public class B{
	static Scanner in;
	static PrintWriter out;
	
	static long INF;
	
	static int n;
	static long p, q, r, ans;
	static long[] array, pref, suff;
	
	public static void main(String args[]){
		in = new Scanner(System.in);
		out = new PrintWriter(System.out);
		
		INF = 8;
		for (int i = 1; i <= 18; i++) INF *= 10;
		
		n = in.nextInt();
		p = in.nextLong();
		q = in.nextLong();
		r = in.nextLong();
		
		array = new long[100005];
		pref = new long[100005];
		suff = new long[100005];
		
		for (int i = 1; i <= n; i++){
			array[i] = in.nextLong();
		}
		
		pref[0] = -INF;
		for (int i = 1; i <= n; i++){
			pref[i] = Math.max(pref[i - 1], p * array[i]);
		}
		
		suff[n + 1] = -INF;
		for (int i = n; i >= 1; i--){
			suff[i] = Math.max(suff[i + 1], r * array[i]);
		}
		
		ans = -INF;
		for (int i = 1; i <= n; i++){
			ans = Math.max(ans, pref[i] + suff[i] + q * array[i]);
		}
		out.println(ans);
		out.close();
	}
}
