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.pb.codeforces.practice;

import java.util.Scanner;

public class CF855B {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		long p = in.nextLong();
		long q = in.nextLong();
		long r = in.nextLong();
		long[] arr = new long[100005];
		for(int i=1; i<=n; i++)
			arr[i] = in.nextLong();
		long[] lmax = new long[100005];
		long[] lmin = new long[100005];
		long[] rmax = new long[100005];
		long[] rmin = new long[100005];
		lmax[1] = arr[1]; lmin[1] = arr[1];
		for(int i=2; i<=n; i++) {
			lmax[i] = Math.max(lmax[i-1], arr[i]);
			lmin[i] = Math.min(lmin[i-1], arr[i]);
		}
		rmax[n] = arr[n]; rmin[n] = arr[n];
		for(int i=n-1; i>=1; i--) {
			rmax[i] = Math.max(rmax[i+1], arr[i]);
			rmin[i] = Math.min(rmin[i+1], arr[i]);
		}
		long ans = Long.MIN_VALUE;
		for(int i=1; i<=n; i++) {
			long mv = arr[i]*q;
			long lv = (p<0 ? lmin[i]*p : lmax[i]*p);
			long rv = (r<0 ? rmin[i]*r : rmax[i]*r);
			ans = Math.max(ans, mv+lv+rv);
		}
		System.out.println(ans);
	}
}
