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: Accepted)

import java.io.InputStream;
import java.io.PrintStream;
import java.util.*;

public class Main {

  public static void main(String[] args) {
    solve(System.in, System.out);
  }

  static void solve(InputStream is, PrintStream os) {
    Scanner sc = new Scanner(is);

    /* read */
    int n = sc.nextInt();
    int k = sc.nextInt();
    int[] a = new int[n];

    for (int i = 0; i < n; i++) {
      a[i] = sc.nextInt();
    }

    SortedSet<Integer> set = new TreeSet<>();

    // find already k-length ordered segment
    Set<Integer> orderedIndexes = new HashSet<>();
    int l = 0;
    int r = 1;
    for (l = 0; l < n; l++) {
      if (l == r) r++;
      while (r < n && a[r - 1] < a[r]) r++;
      if (k == r - l) orderedIndexes.add(l);
    }

    for (int i = 0; i < k; i++) {
      set.add(a[i]);
    }

    int res = n - k + 1;

    for (int i = 0; i < n - k; i++) {
      int min = set.first();
      int max = set.last();
      set.remove(a[i]);
      set.add(a[i + k]);
      if (min == a[i] && max < a[i + k]) res--;
    }

    if (orderedIndexes.size() > 1) {
      res -= orderedIndexes.size() - 1;
    }

    os.println(res);

  }

}