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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;

const int MAX_N=2e5+10;

int P[MAX_N],N,cnt[MAX_N],ans=0,K;
set <int> s;

int main(){
	scanf("%d%d",&N,&K);
	for(int i=0;i<N;i++) scanf("%d",&P[i]);
	for(int i=0;i<N;i++) cnt[i]=0;
	cnt[0]=0;
	for(int i=1;i<N;i++){
		if(P[i]>P[i-1]) cnt[i]=cnt[i-1]+1;
		else cnt[i]=0;
		if(cnt[i]>=K-1) ans=1; 
	}
	for(int i=0;i<K;i++) s.insert(P[i]);
	if(cnt[K-1]!=K-1) ans++;
	for(int i=K;i<N;i++){
		s.insert(P[i]);
		if(*s.begin()!=P[i-K]||*s.rbegin()!=P[i])
			if(cnt[i]<K-1) ans++;
		s.erase(P[i-K]);
	}
	printf("%d\n",ans);
	return 0;
}