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
//Sorting a Segment:

/*
Website for this problem:
https://atcoder.jp/contests/agc038/tasks/agc038_b
*/

/*
Writer of this code:
LHic
*/

#include <bits/stdc++.h>
using namespace std;
const int N=(2e5+1e4);
int n, k;
int a[N];
int go[N];
set<int>ss;
int ans=0;
int i;
int main()
{
	scanf("%d%d",&n,&k);
	for (i=0;i<n;i++)
	{
		scanf("%d",&a[i]);
	}
 	go[0]=0;
	for(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;
		}
	}
	for(i=0;i<k;i++)
	{
		ss.insert(a[i]);
	}	
	if(go[k-1]!=k-1)
	{
		ans++;
	}
	for(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]);
	}
	printf("%d",ans);
	return 0;
}