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<set>
using namespace std;
int n,k,f[2345678],ct,as,is,as2;
set<int> s;
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++)scanf("%d",&f[i]);
	for(int i=2;i<=k;i++)as+=f[i]<f[i-1];
	for(int i=1;i<=k;i++)s.insert(f[i]);
	if(as==0)is=1;else as2=1;
	for(int i=k+1;i<=n;i++)
	{
		int fg=0;
		if(*s.lower_bound(-100)!=f[i-k])fg=1;
		s.erase(f[i-k]);
		s.insert(f[i]);
		if(s.lower_bound(f[i]+1)!=s.end())fg=1;
		as-=f[i-k+1]<f[i-k];
		as+=f[i]<f[i-1];
		if(as==0)is=1;
		else as2+=fg;
	}
	printf("%d\n",as2+is);
}