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.*;
public class scratch_18{


    static class Reader {
        static BufferedReader reader;
        static StringTokenizer tokenizer;

        static void init(InputStream input) {
            reader = new BufferedReader(
                    new InputStreamReader(input));
            tokenizer = new StringTokenizer("");
        }

        static String next() throws IOException {
            while (!tokenizer.hasMoreTokens()) {
                //TODO add check for eof if necessary
                tokenizer = new StringTokenizer(
                        reader.readLine());
            }
            return tokenizer.nextToken();
        }

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

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

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


    public static void main(String[] args)  throws IOException {
        Reader.init(System.in);
        BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));

            int n= Reader.nextInt();
            int x= Reader.nextInt();
            long dp[][][]= new long[n+1][3][3];    // second state is if before max sum subarray(0), included(1), after(2).
            // Third state is for before x(0), included in x(1), after x(2).
            long arr[]= new long[n];
            for (int i = 0; i <n ; i++) {
                arr[i]= Reader.nextLong();
            }

            for (int i = 0; i <n ; i++) {
                // case when maxmimum subarray has not started
                dp[i+1][0][0]= dp[i][0][0];     // max not started, x not started

                dp[i+1][0][1]=Math.max(dp[i+1][0][1],Math.max(dp[i][0][1],dp[i][0][0]));      // max not started, x has started but not ended

                // maxmimum of pichla not in max but x started, pichla not in max and x also not started


                dp[i+1][0][2]= Math.max(dp[i+1][0][2],Math.max(dp[i][0][1],dp[i][0][0]));         // max not started, x has ended.

                // maxmimum of pichla not in max but in x, pichla not in max and neither in x


                // case when maxmimum subarray has started

                dp[i+1][1][0]= Math.max(dp[i+1][1][0],Math.max(dp[i][0][0],dp[i][1][0])+arr[i]);     // max has started, x has not started

                // maximum of pichla not in max and x not started, pichla in max and x not started

                dp[i+1][1][1]= Math.max(dp[i+1][1][0],Math.max(dp[i][1][1],Math.max(dp[i][1][0],Math.max(dp[i][0][1],dp[i][0][0])))+x*arr[i]);       // max has started, x has started.

                // maximum of pichla in max and x, pichla not in max but x, pichla in x but not in max, pichla not in max and not in x

                dp[i+1][1][2]= Math.max(dp[i+1][1][2],Math.max(dp[i][0][1],Math.max(dp[i][0][2],Math.max(dp[i][1][1],dp[i][1][2])))+arr[i]);  // max has started, x has ended

                // maximum of pichla not in max but in x, pichla not in max , not in x, pichla in max and in x, pichla in max but not in x


                // case when maximum subarray has ended

                dp[i+1][2][0]= Math.max(dp[i+1][2][0],Math.max(dp[i][2][0],dp[i][1][0]));   // maximum subarray has ended, x has not started
                // max of either previous in max subarray or not.

                dp[i+1][2][2]= Math.max(dp[i+1][2][2],Math.max(dp[i][1][2],Math.max(dp[i][1][1],Math.max(dp[i][2][1],dp[i][2][2]))));  // maximum has ended, x has ended
                // previous in max and x and all 4 cases

            }
            long max= Long.MIN_VALUE;
        for (int i = 0; i <3 ; i++) {
            for (int j = 0; j <3 ; j++) {
                max= Math.max(max,dp[n][i][j]);
            }
        }
        out.append(max+"\n");


//            out.append(dp[n-1][2][2]+"\n");

        out.flush();
        out.close();
    }
}








