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
#include<bits/stdc++.h>
using namespace std;
int n,k,a[200100],mn[200100],mx[200100];
deque<int> q;
int main(){
	ios::sync_with_stdio(0);
	cin.tie(NULL);
	cout.tie(NULL);
	cin>>n>>k;
	for(int i=1;i<=n;i++) cin>>a[i];
	while(q.size()) q.pop_back();
	for(int i=n;i>=1;i--){
		while(q.size()&&q.front()>=i+k) q.pop_front();
		while(q.size()&&a[q.back()]>a[i]) q.pop_back();
		q.push_back(i);
		if(q.size()==1) mn[i]=1;
	}
	while(q.size()) q.pop_back();
	for(int i=1;i<=n;i++){
		while(q.size()&&q.front()<=i-k) q.pop_front();
		while(q.size()&&a[q.back()]<a[i]) q.pop_back();
		q.push_back(i);
		if(q.size()==1) mx[i]=1;
	}
	int num=0;
	for(int i=1,j=1;i<=n;i=j){
		j++;
		while(j<=n&&a[j]>a[j-1]) j++;
		if(j>=i+k) num++;
	}
	int tot=n-k+1-max(num-1,0);
	for(int i=k;i<=n;i++) tot-=(mx[i]&&mn[i-k]);
	cout<<tot<<endl;
	return 0;
}
