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 <iostream>
#include <cstring>
#include <set>
using namespace std;

const int MAXN = 200010;
int a[MAXN];
int main() {
	int n, k;
	cin>>n>>k;
	for (int i=0; i<n; i++)
		cin>>a[i];
	set<int> s;
	for (int i=0; i<k; i++)
		s.insert(a[i]);
	int ans = 0, flagasc = 0;
	int rcount = 0;
	for (int i=1; i<k; i++)
		if (a[i-1] > a[i])
			rcount++;
	if (rcount)
		ans = 1;
	else
		flagasc = 1;
	for (int i=k; i<n; i++) {
		int round = 0;
		s.erase(a[i-k]);
		if (a[i-k] > a[i-k+1])
			rcount--;

		if (a[i-k] > *s.begin() && rcount != 0)
			round = 1;
		if (a[i] < *s.rbegin())
			round = 1;

		s.insert(a[i]);
		if (a[i-1] > a[i])
			rcount++;
		ans += round;
		if (rcount == 0)
			flagasc = 1;
	}
	cout<<ans + flagasc<<endl;
	return 0;
}