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>
using namespace std;
const int N=400010;
int a[N],k,n,b[N],c[N];
struct Q{
	int id,q[N],l=1,r;
	void ins(int ps){
		if(id){
			while(l<=r&&a[ps]>a[q[r]])r--;
		}
		else{
			while(l<=r&&a[ps]<a[q[r]])r--;
		}
		q[++r]=ps;
		while(l<=r&&q[l]<ps-k)l++;
	}
	int f(){
		return a[q[l]];
	}
}mx,mi;
int main(){
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		a[i]++;
	}
	for(int i=2;i<=n;i++){
		if(a[i]<a[i-1])b[i]=1;
		c[i]=c[i-1]+b[i];
	}
	int tot=0,ans=n-k+1;
	for(int i=k;i<=n;i++){
		if(c[i]==c[i-k+1])tot++;
	}
	if(tot)ans-=tot-1;
	mx.id=1;
	for(int i=1;i<=n;i++){
		mx.ins(i),mi.ins(i);
		if(i>k&&mx.f()==a[i]&&mi.f()==a[i-k]){
			if(c[i]==c[i-k+1]&&c[i-1]==c[i-k])continue;
			ans--;
		}
	}
	printf("%d\n",ans);
}