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.
//package albert.codeforces.marvologauntsring855b;

import java.util.Scanner;

public class Solution {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int p = scan.nextInt();
        int q = scan.nextInt();
        int r = scan.nextInt();
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = scan.nextInt();
        }
       /* long p = 1;
        long q = 2;
        long r = -3;
        int[] arr = {-1, -2, -3, -4, -5};*/
        System.out.println(solve(p, q, r, arr));
    }

    static long solve(long p, long q, long r, int[] arr) {
        int n = arr.length;

        long[] wtf = {p, q, r};

        long[][] dp = new long[n][3];
        dp[0][0] = arr[0] * p;
        for (int i = 1; i < n; i++) {
            dp[i][0] = Math.max(dp[i - 1][0], arr[i] * p);
        }
        for (int i = 1; i < 3; i++) {
            dp[0][i] = dp[0][i - 1] + arr[0] * wtf[i];
            for (int j = 1; j < n; j++) {
                dp[j][i] = Math.max(dp[j - 1][i], arr[j] * wtf[i] + dp[j][i - 1]);
            }
        }
        long max = Long.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            max = Math.max(max, dp[i][2]);
        }
        return max;
    }
}
