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:
//besmellah
#include <bits/stdc++.h>
using namespace std;

const int maxn = 2e5 + 10;
int p[maxn], dp[maxn];
bool B[maxn];
set <int> S;

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n, k;
	cin >> n >> k;
	dp[k - 1] = 1;
	for (int i = 0; i < n; i++)
		cin >> p[i];
	for (int i = 0; i < k; i++)
		S.insert(p[i]);
	int g = n, l = 1;
	for (int i = 0; i < n - 1; i++){
		if (p[i] > p[i + 1])
			l = 1;
		else
			l ++;
		if (l == k){
			g = min(g, i + 1);
		}
		if (l >= k - 1){
			B[i + 1] = true;
		}
	}
	for (int i = k; i < n; i++){
		bool f = *(S.begin()) == p[i - k];
		S.erase(p[i - k]);
		S.insert(p[i]);
		auto x = S.end();
		x --;
		if (((g < i - k && B[i - 1]) || f) && S.find(p[i]) == x)
			dp[i] = dp[i - 1];
		else
			dp[i] = dp[i - 1] + 1;
	}
	cout << dp[n - 1];
}