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 MarveloGauntsRing {
	// http://codeforces.com/problemset/problem/855/B
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		long p = sc.nextInt();
		long q = sc.nextInt();
		long r = sc.nextInt();
		int[] a = new int[n];
		for (int i = 0; i < a.length; i++) {
			a[i] = sc.nextInt();
		}
		sc.close();
		// dp[i][j] = maximum using values up to ai and using j multiplications
		// base dp[i][0] = 0;
		long[][] dp = new long[n][4];
		dp[0][1] = a[0] * p;
		dp[0][2] = dp[0][1] + a[0] * q;
		dp[0][3] = dp[0][2] + a[0] * r;
		for (int i = 1; i < dp.length; i++) {
			dp[i][1] = Math.max(dp[i - 1][1], a[i] * p);
			dp[i][2] = Math.max(dp[i - 1][2], dp[i][1] + a[i] * q);
			dp[i][3] = Math.max(dp[i - 1][3], dp[i][2] + a[i] * r);
		}
		System.out.println(dp[n - 1][3]);
	}
}
