Problem: 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.

Examples

Input

5 3
0 2 1 4 3


Output

2


Input

4 4
0 1 2 3


Output

1


Input

10 4
2 0 1 3 7 5 4 6 8 9


Output

6

Solution (in CPP):

Solution:
#include<iostream>
#include<algorithm>
#include<set>
using lint=int64_t;
using namespace std;

int main()
{
	int N,K;
	int P[200010]={};

	cin >> N >> K;
	for(int i=0;i<N;i++)
		cin >> P[i];
	
	int sum[200010]={};
	for(int i=1;i<N;i++)
		sum[i]=sum[i-1]+(P[i-1]<P[i]?1:0);
	
	int ans=0;
	bool flag=false;
	for(int i=0;i<N-K+1;i++)
	{
		if(sum[i+K-1]-sum[i]==K-1)
			flag=true;
	}
	if(flag)
		ans++;
	
	for(int i=0;i<N-K+1;i++)
	{
		if(sum[i+K-1]-sum[i]!=K-1)
			ans++;
	}

	set<int> s;
	for(int i=0;i<K;i++)
		s.insert(P[i]);
	for(int i=0;i<N-K;i++)
	{
		s.insert(P[i+K]);
		if(sum[i+K-1]-sum[i]!=K-1 && sum[i+K]-sum[i+1]!=K-1 && sum[i+K-1]-sum[i+1]!=K-2)
		{
			auto minitr=s.begin();
			auto maxitr=s.rbegin();
			if(*minitr==P[i] && *maxitr==P[i+K])
				ans--;
		}
		s.erase(P[i]);
	}

	cout << ans << endl;
	return 0;
}