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.nio.CharBuffer;
import java.util.NoSuchElementException;

public class P1155D {

    public static void main(String[] args) {
        SimpleScanner scanner = new SimpleScanner(System.in);
        PrintWriter writer = new PrintWriter(System.out);

        int n = scanner.nextInt();
        int k = scanner.nextInt();

        long[] x = new long[n];
        long ans = 0;
        long[][] f = new long[2][3];
        int cur = 0;
        for (int i = 0; i < n; ++i) {
            cur ^= 1;
            x[i] = scanner.nextLong();
            f[cur][0] = (f[cur ^ 1][0] < 0 ? 0 : f[cur ^ 1][0]) + x[i];
            f[cur][1] = Math.max(f[cur ^ 1][0] < 0 ? 0 : f[cur ^ 1][0], f[cur ^ 1][1] < 0 ? 0 : f[cur ^ 1][1]) + k * x[i];
            f[cur][2] = Math.max(f[cur ^ 1][1] < 0 ? 0 : f[cur ^ 1][1], f[cur ^ 1][2] < 0 ? 0 : f[cur ^ 1][2]) + x[i];
            ans = Math.max(ans, Math.max(f[cur][0], Math.max(f[cur][1], f[cur][2])));
        }
        writer.println(ans);

        writer.close();
    }

    private static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    private static class SimpleScanner {

        private static final int BUFFER_SIZE = 10240;

        private Readable in;
        private CharBuffer buffer;
        private boolean eof;

        SimpleScanner(InputStream in) {
            this.in = new BufferedReader(new InputStreamReader(in));
            buffer = CharBuffer.allocate(BUFFER_SIZE);
            buffer.limit(0);
            eof = false;
        }


        private char read() {
            if (!buffer.hasRemaining()) {
                buffer.clear();
                int n;
                try {
                    n = in.read(buffer);
                } catch (IOException e) {
                    n = -1;
                }
                if (n <= 0) {
                    eof = true;
                    return '\0';
                }
                buffer.flip();
            }
            return buffer.get();
        }

        void checkEof() {
            if (eof)
                throw new NoSuchElementException();
        }

        char nextChar() {
            checkEof();
            char b = read();
            checkEof();
            return b;
        }

        String next() {
            char b;
            do {
                b = read();
                checkEof();
            } while (Character.isWhitespace(b));
            StringBuilder sb = new StringBuilder();
            do {
                sb.append(b);
                b = read();
            } while (!eof && !Character.isWhitespace(b));
            return sb.toString();
        }

        int nextInt() {
            return Integer.valueOf(next());
        }

        long nextLong() {
            return Long.valueOf(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}
