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<cstdio>
int a[200050],b[200050];
struct que1{
	int q[200050],h,r;
	inline void init(){
		h=1;r=0;
	}
	inline void push(int x){
		while(h<=r&&x>q[r])--r;
		q[++r]=x;
	}
	inline void pop(int x){if(x==q[h])++h;}
}Q;
struct que2{
	int q[200050],h,r;
	inline void init(){
		h=1;r=0;
	}
	inline void push(int x){
		while(h<=r&&x<q[r])--r;
		q[++r]=x;
	}
	inline void pop(int x){if(x==q[h])++h;}
}q;
int main(){
	int n,k,i,cnt=1,ans=1;
	bool f;
	scanf("%d%d",&n,&k);
	for(i=1;i<=n;++i)scanf("%d",a+i);
	for(b[1]=1,i=2;i<=n;++i)if(a[i]>a[i-1])b[i]=b[i-1];
	else b[i]=++cnt;
	Q.init();q.init();
	for(i=1;i<=k;++i)Q.push(a[i]),q.push(a[i]);
	f=b[k]==b[1];
	for(;i<=n;++i){
		Q.pop(a[i-k]);q.pop(a[i-k]);
		if(q.q[q.h]<a[i-k]||a[i]<Q.q[Q.h])if(b[i]==b[i-k+1])if(!f){++ans;f=1;}
		else;
		else ++ans;
		Q.push(a[i]);q.push(a[i]);
	}
	printf("%d\n",ans);
	return 0;
}