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.BufferedWriter;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Scanner;

public class B855 {
	static int SIZE = 100010, nodeCount = 0;
	static long[] first = new long[SIZE * 3];
	static long[] third = new long[SIZE * 3];
	
	static void buildTree(int l, int r) {
		int cur = ++nodeCount;
		first[cur] = Long.MIN_VALUE;
		third[cur] = Long.MIN_VALUE;
		if (l != r) {
			int m = (l + r) / 2;
			buildTree(l, m);
			buildTree(m + 1, r);
		}
	}
	static void update(long[] val, int c, int l, int r, int p, long v) {
		if (l == r) {
			val[c] = v;
		} else {
			int m = (l + r) / 2;
			if (p <= m) {
				update(val, 2 * c, l, m, p, v);
			} else {
				update(val, 2 * c + 1, m + 1, r, p, v);
			}
			val[c] = Math.max(val[c * 2], val[c * 2 + 1]);
		}
	}
	static long query(long[] val, int c, int l, int r, int lef, int rig) {
		if (l >= lef && r <= rig) {
			return val[c];
		} else {
			int m = (l + r) / 2;
			long ans = Long.MIN_VALUE;
			if (m >= lef) {
				ans = Math.max(ans, query(val, 2 * c, l, m, lef, Math.min(m, rig)));
			}
			if ((m + 1) <= rig) {
				ans = Math.max(ans, query(val, 2 * c + 1, m + 1, r, Math.max(lef, m+1), rig));
			}
			return ans;
		}
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		int n = in.nextInt();
		long p = in.nextLong(), q = in.nextLong(), r = in.nextLong();
		int[] a = new int[n];
		// buildTree(1, n);
		for (int i=0; i<n; i++) {
			a[i] = in.nextInt();
			update(first, 1, 1, n, i + 1, p * a[i]);
			update(third, 1, 1, n, i + 1, r * a[i]);
		}

		long ans = Long.MIN_VALUE;
		
		for (int i=0; i<n; i++) {
			long left = query(first, 1, 1, n, 1, i + 1);
			long mid = q * a[i];
			long right = query(third, 1, 1, n, i +1, n);
			ans = Math.max(ans, left + mid + right);
			;
		}
		
		out.println(ans);
		
		out.close();
	}

}
