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.PriorityQueue;
import java.util.Scanner;

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-1;i++) 	p[i] = sc.nextInt();
		sc.close();

		int[] mins = new int[n];
		int[] maxs = new int[n];
		//PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());
		PriorityQueue<Integer> pq = new PriorityQueue<> ();

		for(int j=0; j<k; j++) {
			maxs[0] = Math.max(maxs[0], p[j]);
			pq.add(p[j]);
		}

		for(int i=1; i<n-k; i++) {
			int j=i+k-1;
			pq.add(p[j]);
			if(pq.peek()==p[i-1]) {
				pq.poll();
			}
			mins[i] = pq.peek();
			maxs[i] = Math.max(maxs[i-1], p[j]);
		}

		int ans = 1;
		int tmpm=mins[0];
		int tmpM=maxs[0];
		for(int i=1; i<n-k; i++) {
			if((tmpm == mins[i])&&(tmpM == maxs[i])) {
			}else {
				ans++;
			}
			tmpm = mins[i];
			tmpM = maxs[i];
		}

		System.out.println(ans);

	}
}