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<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<set>
#define maxn 200005
using namespace std;
int n,k;
int p[maxn];
set<int> s1,s2;
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++)
	scanf("%d",&p[i]);
	p[0]=-1;
	int cnt=0,flag=1;
	for(int i=1;i<=k;i++)
	{
		s1.insert(p[i]);
		s2.insert(-p[i]);
		if(p[i]>p[i-1])
			cnt++;
		else
			cnt=1;
	}
	if(cnt==k)
		flag=0;	
	int ans=1;
	for(int i=k+1;i<=n;i++)
	{
		if(p[i]>p[i-1])
			cnt++;
		else
			cnt=1;
		s1.erase(p[i-k]);
		s2.erase(-p[i-k]);
		int minn=(*s1.begin());
		int maxx=(-(*s2.begin()));
		if(maxx<p[i] && minn>p[i-k])
		1;
		else
		{
			if(cnt>=k)
			{
				if(flag==1)
				{
					flag=0;
					ans++;
				}
				else
				{
					flag=0;
				}
			}
			else
				ans++;
		}
		s1.insert(p[i]);
		s2.insert(-p[i]);
	}
	printf("%d\n",ans);
}