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.
/**
 * Created by Ariana Herbst on 9/24/17.
 */

import java.util.*;
import java.io.*;

/**
 * Created by Ariana Herbst on 9/24/17
 */
public class cfsasdfB {

    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        StringBuilder sb = new StringBuilder();
        int n = sc.nextInt();
        long[] p = new long[3];
        for (int i = 0; i < 3; i++) {
            p[i] = sc.nextLong();
        }
        long[] nums = new long[n];
        long[] best1 = new long[n];
        for (int i = 0; i < n; i++)
            nums[i] = sc.nextLong();
        for (int i = 0; i < n; i++) {
            best1[i] = (nums[i] * p[0]);
            if (i > 0)
                best1[i] = Math.max(best1[i], best1[i-1]);
        }
        long[] best2 = new long[n];
        for (int i = 0; i < n; i++) {
            long curr = (nums[i] * p[1]);
            best2[i] = curr + best1[i];
            if (i > 0) {
                best2[i] = Math.max(best2[i], best2[i-1]);
            }
        }
        long[] best3 = new long[n];
        for (int i = 0; i < n; i++) {
            long curr = (nums[i] * p[2]);
            best3[i] = curr + best2[i];
            if (i > 0) {
                best3[i] = Math.max(best3[i], best3[i-1]);
            }
        }

        System.out.println(best3[n-1]);
    }


    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(Reader in) {
            br = new BufferedReader(in);
        }

        public FastScanner() {
            this(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String readNextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        int[] readIntArray(int n) {
            int[] a = new int[n];
            for (int idx = 0; idx < n; idx++) {
                a[idx] = nextInt();
            }
            return a;
        }

        long[] readLongArray(int n) {
            long[] a = new long[n];
            for (int idx = 0; idx < n; idx++) {
                a[idx] = nextLong();
            }
            return a;
        }
    }
}
