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>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<ctime>
#include<vector>
#include<set>

using namespace std;

typedef long long ll;
typedef double db;

const int maxn=(int)2e5+9;
const db eps=1e-6;
const int mp=72;

int a[maxn];
int n;
int k;

void init(){
	scanf("%d%d",&n,&k);
	for(int i=0;i<n;i++){
		scanf("%d",&a[i]);
	}
}

int rs[maxn];
set<int> s;

int main(){
	init();
	rs[0]=1;
	int ans=0;
	int mx=0;
	for(int i=1;i<n;i++){
		if(a[i]>a[i-1]){
			rs[i]=rs[i-1]+1;
		}
		else{
			rs[i]=1;
		}
		mx=max(rs[i],mx);
	}
	if(mx>=k){
		ans++;
	}
	int fl=0;
	s.clear();
	for(int i=0;i<k;i++){
		s.insert(a[i]);
		if(*s.rbegin()!=a[i]){
			fl=1;
		}
	}
	if(fl)ans++;
	for(int i=k;i<n;i++){
		int fl=0;
		if(*s.begin()!=a[i-k]){
			fl=1;
		}
		s.erase(a[i-k]);
		s.insert(a[i]);
		if(*s.rbegin()!=a[i]){
			fl=1;
		}
		if(fl&&rs[i]<k)ans++;
	}
	printf("%d\n",ans);
	return 0;
}
