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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class ManthanB {
	static MyScanner in = new MyScanner();
	static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

	public static void main(String[] args) {

		long n = in.nextLong(), p = in.nextLong(), q = in.nextLong(), r = in.nextLong();
		// long n = in.nextLong();
		// long k = in.nextLong();
		// String s = in.next();

		long[] arr = new long[(int) n];
		for (int i = 0; i < n; i++) {
			arr[i] = in.nextLong();
		}
		long[] pref = new long[(int) n];
		pref[0] = arr[0] * p + arr[0] * q;
		long maxSum = Long.MIN_VALUE, maxP = arr[0] * p;
		for (int i = 1; i < n; i++) {
			if (arr[i] * p >= maxP) {
				maxP = arr[i] * p;
			}
			// maxSum = Math.max(maxSum,);
			pref[i] = Math.max(pref[i - 1], maxP + q * arr[i]);
		}
		// System.out.println(Arrays.toString(pref));
		maxSum = Long.MIN_VALUE;
		// long maxo = Integer.MIN_VALUE;
		for (int i = 0; i < n; i++) {
			// maxo = Math.max(maxo, pref[i]);
			maxSum = Math.max(pref[i] + arr[i] * r, maxSum);
		}
		out.println(maxSum);
		/*
		 * Dont delete
		 */
		out.close();
		/*
		 * 
		 */
	}

	// BIN seach
	// while (l != r) {
	// mid = (l + r) / 2;
	// if (check(mid)) {
	// ans = mid;
	// l = mid + 1;
	// } else {
	// r = mid;
	// }
	// }
	public static long GCD(long a, long b) {
		return b == 0 ? a : GCD(b, a % b);
	}

	public static long LCM(long a, long b) {
		return a * b / GCD(a, b);
	}

	public static class MyScanner {
		BufferedReader br;
		StringTokenizer st;

		public MyScanner() {
			br = new BufferedReader(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 nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
}
