You are given an array a consisting of n integers. Beauty of array is the maximum sum of some consecutive subarray of this array (this subarray may be empty). For example, the beauty of the array [10, -5, 10, -4, 1] is 15, and the beauty of the array [-3, -5, -1] is 0.

You may choose at most one consecutive subarray of a and multiply all values contained in this subarray by x. You want to maximize the beauty of array after applying at most one such operation.

Input

The first line contains two integers n and x (1 ≤ n ≤ 3 ⋅ 10^5, -100 ≤ x ≤ 100) — the length of array a and the integer x respectively.

The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the array a.

Output

Print one integer — the maximum possible beauty of array a after multiplying all values belonging to some consecutive subarray x.

Examples

Input


5 -2
-3 8 -2 1 -6


Output


22


Input


12 -3
1 3 3 7 1 3 3 7 1 3 3 7


Output


42


Input


5 10
-1 -2 -3 -4 -5


Output


0

Note

In the first test case we need to multiply the subarray [-2, 1, -6], and the array becomes [-3, 8, 4, -2, 12] with beauty 22 ([-3, 8, 4, -2, 12]).

In the second test case we don't need to multiply any subarray at all.

In the third test case no matter which subarray we multiply, the beauty of array will be equal to 0.
import java.io.*;
import java.util.Scanner;

/**
 * @author <a href="https://github.com/Klnsyf-Sun">Klnsyf-Sun</a>
 */
public class Main {

    private static void solve() throws IOException {
        int n = nextInt(), x = nextInt();
        long[] a = new long[n];
        for (int index = 0; index < n; index++) {
            a[index] = nextInt();
        }
        long[] sum = new long[n], maxPrefix = new long[n], maxSuffix = new long[n];
        for (int index = 0; index < n; index++) {
            sum[index] = a[index] + (index == 0 ? 0 : sum[index - 1]);
            maxPrefix[index] = Math.max(0, a[index] + (index == 0 ? 0 : maxPrefix[index - 1]));
        }
        for (int index = n - 1; index >= 0; index--) {
            maxSuffix[index] = Math.max(0, a[index] + (index == n - 1 ? 0 : maxSuffix[index + 1]));
        }
        long maxBeauty = 0;
        /*
        maxBeauty
        = maxPrefix[leftBound - 1] + (sum[rightBound] - sum[leftBound - 1]) * x + maxSuffix[rightBound + 1]
        = (maxPrefix[leftBound - 1] - sum[leftBound - 1] * x)    <- maxLeftBound
        + (sum[rightBound] * x + maxSuffix[rightBound + 1])      <- maxRightBound
         */
        long maxLeftBound = 0;
        for (int index = -1; index < n; index++) {  // index: rightBound
            maxBeauty = Math.max(maxBeauty, maxLeftBound + (index == -1 ? 0 : sum[index]) * x + (index == n - 1 ? 0 : maxSuffix[index + 1]));
            maxLeftBound = Math.max(maxLeftBound, (index == -1 ? 0 : maxPrefix[index]) - (index == -1 ? 0 : sum[index]) * x);
        }
        out.println(maxBeauty);
    }


    private static StreamTokenizer in;
    private static Scanner sc;
    private static PrintWriter out;

    static {
        in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        in.ordinaryChars(33, 126);
        in.wordChars(33, 126);
        sc = new Scanner(new BufferedInputStream(System.in));
        out = new PrintWriter(new OutputStreamWriter(System.out));
    }

    public static void main(String[] args) throws IOException {
        solve();
        sc.close();
        out.flush();
    }

    private static int nextInt() throws IOException {
        return Integer.parseInt(nextString());
    }

    private static long nextLong() throws IOException {
        return Long.parseLong(nextString());
    }

    private static double nextDouble() throws IOException {
        return Double.parseDouble(nextString());
    }

    private static String nextString() throws IOException {
        in.nextToken();
        return in.sval;
    }

    private static boolean hasNext() throws IOException {
        int next = in.nextToken();
        in.pushBack();
        return next != StreamTokenizer.TT_EOF;
    }
}