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;
typedef pair<int,int>pii;
const int N=2e5+5;
int n,k,p[N],i,s[N],ans;
bool b[N],bb[N];
struct Q{
	pii q[N];int t,w,O;
	inline void ins(pii u){
		for(;t<=w && (q[w]<u)==O;--w);
		q[++w]=u;
	}
	inline int ask(int x){
		for(;q[t].second<x;)++t;
		return q[t].first;
	}
}q0,q1;
int main(){
	scanf("%d%d",&n,&k);
	for(i=1;i<=n;++i)scanf("%d",p+i);
	for(i=1;i<=n;++i)s[i]=s[i-1]+(p[i]>p[i+1]);
	q0.O=0;q1.O=1;q0.t=q1.t=1;
	for(i=2;i<n;++i){
		q0.ins(pii(p[i],i)),q1.ins(pii(p[i],i));
		if(i>=k){
			if(q0.ask(i-k+2)>p[i-k+1]){
				if(q1.ask(i-k+2)<p[i+1])
					b[i+1]=1;
			}
		}
	}
	for(i=k;i<=n;++i)ans+=bb[i]=s[i-1]-s[i-k]>0;
	for(i=k+1;i<=n;++i)ans-=bb[i-1] && bb[i] && b[i];
	for(i=k;i<=n;++i)if(!bb[i]){ans++;break;}
	printf("%d\n",ans);
	return 0;
}