Problem:
Score : 700 points

Problem Statement

Snuke has a permutation (P_0,P_1,\cdots,P_{N-1}) of (0,1,\cdots,N-1).

Now, he will perform the following operation exactly once:

Choose K consecutive elements in P and sort them in ascending order.

Find the number of permutations that can be produced as P after the operation.

Constraints

2 \leq N \leq 200000

2 \leq K \leq N

0 \leq P_i \leq N-1

P_0,P_1,\cdots,P_{N-1} are all different.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
P_0 P_1 \cdots P_{N-1}

Output

Print the number of permutations that can be produced as P after the operation.

Sample Input 1

5 3
0 2 1 4 3

Sample Output 1

2

Two permutations can be produced as P after the operation: (0,1,2,4,3) and (0,2,1,3,4).

Sample Input 2

4 4
0 1 2 3

Sample Output 2

1

Sample Input 3

10 4
2 0 1 3 7 5 4 6 8 9

Sample Output 3

6

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.*;
import java.util.*;

class Main {
    static Scanner scanner = new Scanner();
    static long mod = 1000000007;

    public static void main(String[]$) {
        int n = scanner.nextInt();
        int k = scanner.nextInt();
        int[] a = new int[n];
        TreeSet<Integer> set = new TreeSet<>();
        for (int i = 0; i < n; i++) {
            a[i] = scanner.nextInt();
            if (i <= k) {
                set.add(a[i]);
            }
        }
        int count = 0;
        int c = 0;
        for (int i = 1; i < n; i++) {
            if (a[i - 1] < a[i]) {
                c++;
            } else {
                if (c >= k - 1) {
                    count++;
                }
                c = 0;
            }
        }
        if (c >= k - 1) {
            count++;
        }
        if (count > 0) {
            count--;
        }
        if (n != k) {
            for (int i = k + 1; i < n; i++) {
                if (set.first().equals(a[i - k - 1]) && set.last().equals(a[i - 1])) {
                    count++;
                }
                set.remove(a[i - k - 1]);
                set.add(a[i]);
            }
        }
        System.out.println(n - k + 1 - count);
    }

    static class Scanner {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in), 1 << 15);
        StringTokenizer tokenizer;

        String next() {
            try {
                while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                    tokenizer = new StringTokenizer(reader.readLine());
                }
            } catch (IOException ignored) {
            }
            return tokenizer.nextToken();
        }

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

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