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.util.*;
 
public class Main {
    public static void main(String[] args){
      Scanner sc = new Scanner(System.in);
      int N = sc.nextInt();
      int K = sc.nextInt();
      
      int P[] = new int[N];
        
      for (int i = 0; i < N; i++) {
        P[i] = sc.nextInt();
      }
      
      int same_count = 0;
      int motionless_count = 0;
      boolean order = true;
      boolean last = false;
      
      for (int i = 0; i < N - K + 1; i++) {
        order = true;
        
        if (i == N - K) {
          for (int j = 0; j < K - 1; j++) {
            if (P[i + j] > P[i + j + 1]) {
              break;
            }
            if (j == K - 2) {           
              motionless_count += 1;
            }
            if (last) {
              same_count -= 1;
            }
          }
          break;
        }
        
        for (int j = 0; j < K; j++) {
          if (order && j < K - 1 && P[i + j] > P[i + j + 1]) {
            order = false;
          }
          if (order && j == K - 2) {
            motionless_count += 1;
          }
          
          if (P[i] > P[i + j + 1]) {
            break;
          }
          if (P[i + K] < P[i + j]) {
            break;
          }
          if (!order && j == K - 1) {
            same_count += 1;
            if (i == N - K - 1) {
              last = true;
            }
          }       
        }
      }
      if (motionless_count > 1) {
        motionless_count -= 1;
      } else {
        motionless_count = 0;
      }
      System.out.print(N - K + 1 - same_count - motionless_count);
    }
}