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>

const int N=200005;

int n,k,a[N],s[N],mn[N],mx[N],pre[N];

int main()
{
	scanf("%d%d",&n,&k);
	for (int i=1;i<=n;i++) scanf("%d",&a[i]);
	int h=1,t=0;
	for (int i=1;i<=n;i++)
	{
		while (h<=t&&a[s[t]]>a[i]) t--;
		if (h<=t&&i-s[h]>=k) h++;
		s[++t]=i;
		mn[i]=s[h];
	}
	h=1,t=0;
	for (int i=1;i<=n;i++)
	{
		while (h<=t&&a[s[t]]<a[i]) t--;
		if (h<=t&&i-s[h]>=k) h++;
		s[++t]=i;
		mx[i]=s[h];
	}
	for (int i=1;i<n;i++) pre[i]=pre[i-1]+(a[i]<a[i+1]);
	int ans=n-k+1;
	bool flag=0;
	for (int i=1;i+k-1<=n;i++)
	{
		if (pre[i+k-2]-pre[i-1]==k-1) ans+=flag^1,flag=1;
		if (pre[i+k-2]-pre[i-1]==k-1||i>1&&mn[i+k-2]==i-1&&mx[i+k-1]==i+k-1) ans--;
	}
	printf("%d\n",ans);
	return 0;
}