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.io.PrintWriter;
import java.util.Scanner;

public class B {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int n = in.nextInt();
        long p = in.nextLong();
        long q = in.nextLong();
        long r = in.nextLong();
        long[] a = new long[n];

        for (int i = 0; i < n; i++) {
            a[i] = in.nextLong();
        }


        long[] min0 = new long[n];
        long[] max0 = new long[n];
        min0[0] = a[0];
        max0[0] = a[0];

        for (int i = 1; i < n; i++) {
            min0[i] = Math.min(a[i], min0[i - 1]);
            max0[i] = Math.max(a[i], max0[i - 1]);
        }

        long[] minN = new long[n];
        long[] maxN = new long[n];
        minN[n - 1] = a[n - 1];
        maxN[n - 1] = a[n - 1];

        for (int i = n - 2; i >= 0; i--) {
            minN[i] = Math.min(a[i], minN[i + 1]);
            maxN[i] = Math.max(a[i], maxN[i + 1]);
        }


        long[] mr = new long[n];

        for (int i = 0; i < n; i++) {
            mr[i] = r > 0 ? r * maxN[i] : r * minN[i];
        }

        long[] mp = new long[n];
        mp[0] = p * a[0];

        for (int i = 1; i < n; i++) {
            mp[i] = p > 0 ? p * max0[i] : p * min0[i];
        }

        long m = p * a[0] + q * a[0] + r * a[0];

        for (int i = 0; i < n; i++) {
            long s = mp[i] + q * a[i] + mr[i];

            if (s > m) {
                m = s;
            }
        }

        out.println(m);

        out.flush();
    }
}
