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 <vector>
#include <set>
#include <queue>
#include <algorithm>
#include <functional>
using namespace std;

int arr[200000];
set<int> s;

int main()
{
	int n, k;
	int ans = 1;
	int cnt = 0;

	scanf("%d %d", &n, &k);

	for (int i = 0; i < n; i++)
		scanf("%d", &arr[i]);

	int now = 1;
	for (int i = 1; i < n; i++)
	{
		if (arr[i] > arr[i - 1])
			now++;
		else
		{
			if (now >= k)
				cnt++;
			now = 1;
		}
	}

	if (now >= k)
		cnt++;

	if (n == k)
	{
		printf("1");
		return 0;
	}

	for (int i = 0; i < k; i++)
		s.insert(arr[i]);

	for (int i = k; i < n; i++)
	{
		if (arr[i - k] != *s.begin() || arr[i] < *s.rbegin())
			ans++;

		s.erase(arr[i - k]);
		s.insert(arr[i]);
	}

	if (cnt == 0)
		printf("%d", ans);
	else
		printf("%d", ans - cnt + 1);

	return 0;
}