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.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        // 整数の入力
        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 sum=0;
        for(int i=0;i<N;i++) {
            for (int j = i; j < Math.min(i+K, N); j++) {
                if(P[i] > P[j]){
                    sum ++;
                }
            }
        }
        if(sum == 0){sum++;}
        System.out.println(sum);
        // 出力
//        System.out.println((a + b + c) + " " + s);
    }
}