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;
set<int> S;
int a[200010];
int main() {
	int n, k;
	scanf("%d%d", &n, &k);
	for(int i = 1; i <= n; ++ i) {
		scanf("%d", &a[i]);
	}
	int Ans = 1, pp = 0;
	for(int i = 1; i <= k; ++ i) {
		S.insert(a[i]);
		if(a[i] < a[i - 1] && i > 1) {
			++ pp;
		}
	}
	int wt = 0;
	if(pp == 0) Ans = 0, wt = 1;
	//cerr << pp << endl;
	
	for(int i = k + 1; i <= n; ++ i) {
		int cnt = 0;
		if(a[i - k] > a[i - k + 1]) -- pp;
		if(a[i - 1] > a[i]) ++ pp;
		if(pp == 0) {
			S.erase(a[i - k]);
			S.insert(a[i]);
			wt = 1;
			continue;
		}
		if(a[i - k] == *S.begin()) {
			++ cnt;
		}
		S.erase(a[i - k]);
		S.insert(a[i]);
		if(a[i] == *S.rbegin()) {
			++ cnt; 
		}
		if(cnt != 2) ++ Ans; 
	}
	printf("%d\n", Ans + wt);
}