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>

int ri() {
	int n;
	scanf("%d", &n);
	return n;
}


int main() {
	int n = ri();
	int k = ri();
	int a[n];
	for (int i = 0; i < n; i++) a[i] = ri();
	std::set<int> cur;
	for (int i = 0; i < k; i++) cur.insert(a[i]);
	int inc[n];
	for (int i = 0; i < n - 1; i++) inc[i + 1] = a[i] < a[i + 1];
	for (int i = 0; i < n - 1; i++) inc[i + 1] += inc[i];
	int res = 1;
	int tona = 0;
	for (int i = 0; i < n - k; i++) {
		if (*cur.begin() != a[i] || *std::prev(cur.end()) > a[i + k]) res++;
		if (inc[i + k - 1] - inc[i] == k - 1 && inc[i + k] - inc[i + 1] == k - 1) tona++;
		cur.erase(a[i]);
		cur.insert(a[i + k]);
	}
	
	// sorted phase
	int inc_num = 0;
	for (int i = 0; i < n - k + 1; i++) {
		if (inc[i + k - 1] - inc[i] == k - 1) inc_num++;
	}
	if (inc_num > 1) {
		res -= inc_num - 1;
		res += tona;
	}
	std::cout << res << std::endl;
	return 0;
}
