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>
using namespace std;
const int N = 1e6 + 7, mod = 1e9 + 7;
#define debug(x) cerr << #x << " " << x << "\n";
int t[N];
int rep[N];

int re(int x) {
	if(rep[x] != x) rep[x] = re(rep[x]);
	return rep[x];
}

void uni(int a, int b) {
	rep[re(a)] = re(b);
}

int32_t main() {
	ios_base::sync_with_stdio(0);
	cin.tie();
	int n, k;
	cin >> n >> k;
	set <int> S;
	for(int i = 0; i <= n; i++) rep[i] = i;

	for(int i = 0; i < n; i++) {
		cin >> t[i];
	}

	for(int i = 0; i < n; i++) {
		S.insert(t[i]);
		if(i >= k + 1) S.erase(S.find(t[i - k -1]));
		if((*S.begin()) == t[i - k] && (*prev(S.end())) == t[i]) uni(i - 1, i);
	}

	int ost = 0;
	for(int i = 1; i < n; i++) {
		if(t[i] < t[i - 1]) ost = i;
		if(i - ost >= k - 1) uni(i, 0);
	}

	set <int> ss;
	for(int i = k - 1; i < n; i++) {
		ss.insert(re(i));
	}

	cout << ss.size();
}