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 MAX_N=2e5+10;

deque <int> Q1,Q2;
int N,K,P[MAX_N],cnt[MAX_N],ans=0;

int main(){
	scanf("%d%d",&N,&K);
	for(int i=0;i<N;i++) scanf("%d",&P[i]);
	cnt[0]=0;
	for(int i=1;i<N;i++){
		if(P[i]>P[i-1]) cnt[i]=cnt[i-1]+1;
		else cnt[i]=0;
		if(cnt[i]>=K-1) ans=1;
	}
	if(cnt[K-1]!=K-1) ans++;
	for(int i=0;i<N;i++){
		while(Q1.size()&&P[Q1.back()]>P[i]) Q1.pop_back();
		Q1.push_back(i);
		while(Q2.size()&&P[Q2.back()]<P[i]) Q2.pop_back();
		Q2.push_back(i);
		if(i>=K){
			if(Q1.front()!=i-K||Q2.front()!=i)
				if(cnt[i]<K-1) ans++;
			if(Q1.front()==i-K) Q1.pop_front();
			if(Q2.front()==i-K) Q2.pop_front();
		}	
	}
	printf("%d\n",ans);
	return 0;
}