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:
#include <bits/stdc++.h>
using namespace std;
int n, m, a[200500], mx, flag, Ans;
int b[200500], c[200500], bo[200500], num;
set<int> ss;
int main(){
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i++)
		scanf("%d", &a[i]);
	for(int i = 2; i <= n; i++){
		b[i] = a[i] > a[i - 1] ? 1 : 0;
		c[i] = b[i] ? c[i - 1] + 1 : 0;
		if(c[i] >= m - 1){ 
			num++;
			bo[i] = 1;
		}
	}
	if(num) Ans = 1;

	for(int i = 1; i <= m; i++){
		if(a[i] < mx) flag = 1;
		else mx = a[i];
		ss.insert(a[i]);
	}
	
	if(flag) Ans++;

	for(int j = m + 1; j <= n; j++){
		if(a[j - m] == *(ss.begin()) && a[j] > *(--ss.end())){
			0;
		}
		else if(!bo[j]) Ans++;
		ss.erase(a[j - m]);
		ss.insert(a[j]);
	}
	printf("%d\n", Ans);
	return 0;
}