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<bits/stdc++.h>
using namespace std;
const int maxn=200100;
int n,k,a[maxn],fv[maxn],Log2[maxn];
int mnv[maxn][20],mxv[maxn][20];
inline int Qmnv(int l,int r){int t=Log2[r-l+1];return min(mnv[l][t],mnv[r-(1<<t)+1][t]);}
inline int Qmxv(int l,int r){int t=Log2[r-l+1];return max(mxv[l][t],mxv[r-(1<<t)+1][t]);}
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++) scanf("%d",a+i);
	for(int i=2;i<=n;i++) Log2[i]=Log2[i>>1]+1;
	for(int i=2;i<=n;i++) fv[i]=fv[i-1]+(a[i-1]>a[i]),mnv[i][0]=mxv[i][0]=a[i];
	for(int j=1;j<20;j++) for(int i=1;i+(1<<j)-1<=n;i++)
	{
		mnv[i][j]=min(mnv[i][j-1],mnv[i+(1<<(j-1))][j-1]);
		mxv[i][j]=max(mxv[i][j-1],mxv[i+(1<<(j-1))][j-1]);
	}
	int ans=0;
	for(int i=k;i<=n;i++) if(fv[i]-fv[i-k+1]==0){ans=1;break;}
	int lstv=0;
	for(int i=k;i<=n;i++) if(fv[i]-fv[i-k+1])
	{
		if(lstv==i-1&&a[i-k]<Qmnv(i-k+1,i-1)&&Qmxv(i-k+1,i-1)<a[i]) ans--;
		lstv=i;ans++;
	}
	cout<<ans<<endl;
}