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 <stdio.h>
#include <algorithm>
#include <set>
using namespace std;
using ll = long long;
int n, k, p[200010], ord[200010];
set<int> s;
int main() {
	scanf("%d %d", &n, &k);
	for (int i = 0; i < n; i++)
		scanf("%d", p + i);
	for (int i = 0; i < k; i++)
		s.emplace(p[i]);
	for (int i = n - 2; 0 <= i; i--) {
		if (p[i] < p[i + 1])
			ord[i] = ord[i + 1] + 1;
		else
			ord[i] = 0;
	}

	int cnt = n - k + 1, unc = k - 1 <= ord[0];

	for (int i = 0; i + k < n; i++) {
		bool valid = true;
		valid &= (*s.begin()) == p[i];
		s.erase(p[i]);
		s.emplace(p[i + k]);
		valid &= (*s.rbegin()) == p[i + k];

		bool isunc = k - 1 <= ord[i + 1];
		if (1 <= unc && isunc)
			valid = true;
		if (isunc)
			unc = 1;

		if (valid)
			cnt--;
	}
	printf("%d\n", cnt);
}