Problem: 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

Solution (in CPP):

Solution:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<deque>
using namespace std;

const int M = 1000001;
int B,n,m,k,a[M],res,f[M];
deque<int>mn,mx; 
int main()
{
	scanf("%d%d",&n,&k);
	a[0]=0x3f3f3f3f;
	for(int i=1;i<=n;i++) 
	{
		scanf("%d",&a[i]);
		if(a[i]>a[i-1]) f[i]=f[i-1]+1;
	}
	for(int i=1;i<=n;i++)
	{
		while(mn.size() && a[i]<a[mn.back()]) mn.pop_back();
		while(mx.size() && a[i]>a[mx.back()]) mx.pop_back();
		mn.push_back(i), mx.push_back(i);
		if(i<k) continue; res+=1;
		
		if(mn.size() && mn.front()<i-k) mn.pop_front();
		if(mx.size() && mx.front()<i-k) mx.pop_front();
		if(f[i]>=k-1) 
		{
			B+=1; continue;
		}
		if(mx.front()==i && mn.front()==i-k) res-=1;
	}
	if(B) res-=B-1;
	printf("%d\n",res);
}