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.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {

    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        long n = Long.parseLong(st.nextToken());
        long p = Long.parseLong(st.nextToken());
        long q = Long.parseLong(st.nextToken());
        long r = Long.parseLong(st.nextToken());

        long[] ar = new long[(int) n];
        st = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            ar[i] = Long.parseLong(st.nextToken());
        }

        long[][] dp = new long[3][(int) n + 1];

        dp[0][0] = Long.MIN_VALUE;
        dp[1][0] = Long.MIN_VALUE;
        dp[2][0] = Long.MIN_VALUE;

        for (int i = 1; i <= n; i++) {
            dp[0][i] = Math.max(dp[0][i - 1], p * ar[i - 1]);
        }

        for (int i = 1; i <= n; i++) {
            dp[1][i] = Math.max(dp[1][i - 1], dp[0][i] + q * ar[i - 1]);
        }

        for (int i = 1; i <= n; i++) {
            dp[2][i] = Math.max(dp[2][i - 1], dp[1][i] + r * ar[i - 1]);
        }

        System.out.println(dp[2][(int) n]);
    }
}