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 N=2e5+5;

int arr[N];
int issorted[N];
int n,k;
set<int> ss;

int issort(int x)
{
	if(x<k-1) return 0;
	return (issorted[x]-issorted[x-k+1])==k-1;
}

int main()
{
	cin >> n >> k;
	for(int i=0;i<n;i++) {cin >> arr[i]; if(i) issorted[i]=(arr[i]>arr[i-1])+issorted[i-1];}
	for(int i=0;i<k;i++) ss.insert(arr[i]);
	int ans=1-issort(k-1);
	for(int i=k;i<n;i++)
	{
		ss.insert(arr[i]);
		int mn=(*ss.begin()),mx=*(--ss.end());
		//if(issort(i)) cout << i << " " << (mn==arr[i-k]&&mx==arr[i]) << endl;
		if((mn==arr[i-k]&&mx==arr[i]) || issort(i));
		else ans++;
		ss.erase(arr[i-k]);
	}
	for(int i=k-1;i<n;i++) if(issort(i)) {ans++; break;}
	printf("%d\n",ans);
}