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;
const int N=210000;
int n,k;
int a[N];
int go[N];
int main()
{
	cin>>n>>k;
	for(int i=0;i<n;i++) cin>>a[i];
	go[0]=0;
	int ans=0;
	for(int i=1;i<n;i++)
	{
		if(a[i]>a[i-1]) go[i]=go[i-1]+1;
		else go[i]=0;
		if(go[i]>=k-1) ans=1;
	}
	set<int> ss;
	for(int i=0;i<k;i++) ss.insert(a[i]);
	if(go[k-1]!=k-1) ans++;
	for(int i=k;i<n;i++)
	{
		ss.insert(a[i]);
		if(*ss.begin()!=a[i-k] || *ss.rbegin()!=a[i])
		{
			if(go[i]<k-1) ans++;
		}
		ss.erase(a[i-k]);
	}
	cout<<ans<<"\n";
	return 0;
} 