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
#pragma GCC optimize("Ofast",3,"inline")
#include<bits/stdc++.h>
using namespace std;
set<int>sett;
int n,k;
int a[200005];
int x[200005]; 
int ans=0;
int flag;
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	cin>>n>>k;
	for(int i=0;i<n;i++){
		cin>>a[i];
		if(i==0)x[0]=0;
		else{
			if(a[i]>a[i-1])x[i]=x[i-1]+1;
			else x[i]=0;
		}
		if(x[i]>=k-1)flag=1;
	}
	for(int i=0;i<k;i++)sett.insert(a[i]);
	if(x[k-1]<k-1)ans++;
	for(int i=k;i<n;i++){
		sett.insert(a[i]);
		if((a[i]!=*sett.rbegin()||a[i-k]!=*sett.begin())&&x[i]<k-1)ans++;
		sett.erase(a[i-k]); 
	}
	cout<<(ans+flag);
	return 0;
}