Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

bool bo[200010],fl;
int n,k,p[200010],ans,nxt[200010];
set<int> s;

int main()
{
	scanf("%d%d",&n,&k),s.clear(),ans=0,fl=0;
	for (int i=1 ;i<=n; i++) scanf("%d",&p[i]);
	for (int i=1; i<=k; i++) s.insert(p[i]);
	for (int i=1 ;i<n; i++) nxt[i]=(p[i]<p[i+1]);
	nxt[n]=n;
	for (int i=n-1; i; i--) if (nxt[i]) nxt[i]=nxt[i+1]; else nxt[i]=i;
	for (int i=1; i<=n-k+1; i++) bo[i]=(nxt[i]>=i+k-1);
	for (int i=k+1; i<=n; i++)
	{
		bool bo1=0,bo2=0;
		if ((*s.begin())==p[i-k]) bo1=1;
		s.erase(p[i-k]),s.insert(p[i]);
		if ((*s.rbegin())==p[i]) bo2=1;
		if (bo1&&bo2) ans++; else
		if (bo[i-k]) {if (!fl) fl=1; else ans++;}
	}
	if (bo[n-k+1]) {if (!fl) fl=1; else ans++;}
	printf("%d\n",n-k+1-ans);
	return 0;
}