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 com.himanshu.practice.july10;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

/**
 * Created by himanshubhardwaj on 10/07/18.
 */
public class MarvoloGauntSRing {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] s = br.readLine().split(" ");

        int n = (int) Long.parseLong(s[0]);
        long p = Long.parseLong(s[1]);
        long q = Long.parseLong(s[2]);
        long r = Long.parseLong(s[3]);
        long a[] = new long[n];
        long R[] = new long[n];
        long Q[] = new long[n];
        long P[] = new long[n];

        s = br.readLine().split(" ");
        //System.out.prlongln("p=" + p + "\t" + "q=" + q + "\t" + "r=" + r);

        for (int i = 0; i < n; i++) {
            a[i] = Long.parseLong(s[i]);
            R[i] = r * a[i];
            Q[i] = q * a[i];
            P[i] = p * a[i];
            //System.out.prlongln(P[i] + "\t" + Q[i] + "\t" + R[i]);
        }
        //System.out.prlongln();
        //System.out.prlongln();

        for (int i = n - 1; i >= 0; i--) {
            if (i <= (n - 2)) {
                R[i] = Math.max(R[i], R[i + 1]);
            }
            Q[i] = Q[i] + R[i];
            if (i <= (n - 2)) {
                Q[i] = Math.max(Q[i], Q[i + 1]);
            }
            P[i] = P[i] + Q[i];
            if (i <= (n - 2)) {
                P[i] = Math.max(P[i], P[i + 1]);
            }
        }

        System.out.print(P[0]);
//        System.out.println();
//
//        for (int i = 0; i < n; i++) {
//            System.out.print(R[i] + "\t");
//        }
//        System.out.println();
//        for (int i = 0; i < n; i++) {
//            System.out.print(Q[i] + "\t");
//        }
//        System.out.println();
//        for (int i = 0; i < n; i++) {
//            System.out.print(P[i] + "\t");
//        }
//        System.out.println();
    }
}
