Problem: 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

Solution (in CPP):

Solution:
// B.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <set>
#include <vector>

using namespace std;

int main(int argc, char* argv[]) {
	while (true) {
		int n = -1, k = -1;
		cin >> n >> k;
		if (n <= 0 || k <= 0) break;
		vector<int> p(n + 1, n), conseq(n);
		for (int i = 0; i < n; ++i) {
			cin >> p[i];
		}
		int cnt = 0, ans = 0;
		for (int i = n - 1; i >= 0; --i) {
			if (p[i] >= p[i + 1]) {
				cnt = 0;
			}
			conseq[i] = ++cnt;
			if (conseq[i] >= k) {
				ans = 1;
			}
		}
		set<int> s(p.begin(), p.begin() + k);
		ans += conseq[0] < k;
		for (int i = k; i < n; ++i) {
			int low = *(s.begin());
			s.erase(p[i - k]);
			int high = *(s.rbegin());
			if ((low != p[i - k]) || (p[i] < high)) {
				ans += conseq[i - k + 1] < k;
			}
			s.insert(p[i]);
		}
		cout << ans << endl;
	}
	return 0;
}