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;

#define N 300000

int n,m,ans,num,fl,mn[N],mx[N],a[N],stk[N];

int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;++i){
		scanf("%d",a+i);
	}
	for (int i=1,tp=0;i<=n;++i){
		for (;tp&&a[i]>a[stk[tp]];--tp);
		if (!tp||stk[tp]<=i-m) mx[i]=1;
		stk[++tp]=i;
	}
	for (int i=n,tp=0;i;--i){
		for (;tp&&a[i]<a[stk[tp]];--tp);
		if (!tp||stk[tp]>=i+m) mn[i]=1;
		stk[++tp]=i;
	}
	for (int i=1;i<m;++i){
		num+=a[i]<a[i-1];
	}
	for (int i=m;i<=n;++i){
		num+=a[i]<a[i-1];
		if (!num) fl=1;
		else
			if (i==m||!mx[i]||!mn[i-m]) ++ans;
		num-=a[i-m+2]<a[i-m+1];
	}
	ans+=fl;
	printf("%d\n",ans);
	
	return 0;
}
			