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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.StringTokenizer;

public class J {

    final long INF = 1_000_000_000_000_000_000L;

    void solve() {
        int n = readInt();
        int p = readInt();
        int q = readInt();
        int r = readInt();
        long[] a = new long[n];
        for(int i = 0;i<n;i++){
            a[i] = readInt();
        }

        long[][] dp = new long[3][n];
        dp[0][0] = a[0] * p;
        dp[1][0] = a[0] * q + dp[0][0];
        dp[2][0] = a[0] * r + dp[1][0];

        for(int i = 1;i<n;i++){
            dp[0][i] = Math.max(dp[0][i - 1], a[i] * p);
        }
        for(int i = 1;i<n;i++){
            dp[1][i] = Math.max(dp[1][i - 1], dp[0][i] + a[i] * q);
        }
        for(int i = 1;i<n;i++){
            dp[2][i] = Math.max(dp[2][i - 1], dp[1][i] + a[i] * r);
        }
        out.print(dp[2][n - 1]);
    }

    public static void main(String[] args) {
        new J().run();
    }

    private void run() {
        try {
            init();
            solve();
            out.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(-1);
        }
    }

    private BufferedReader in;
    private StringTokenizer tok = new StringTokenizer("");

    private PrintWriter out;

    private void init() {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);

//        try {
//            in = new BufferedReader(new FileReader("absum.in"));
//            out = new PrintWriter(new File("absum.out"));
//        } catch (FileNotFoundException e) {
//            e.printStackTrace();
//        }
    }

    private String readLine() {
        try {
            return in.readLine();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
    }

    private String readString() {
        while (!tok.hasMoreTokens()) {
            String nextLine = readLine();
            if (nextLine == null) return null;
            tok = new StringTokenizer(nextLine);
        }

        return tok.nextToken();
    }

    private int readInt() {
        return Integer.parseInt(readString());
    }

    private long readLong() {
        return Long.parseLong(readString());
    }
}