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 = 2e5 + 10;
int n,k,p[N];
int ismin[N];
int fa[N];
set<int> s;
int gf(int x) {
	return fa[x] == 0 ? x : fa[x] = gf(fa[x]);
}
void merge(int a,int b) {
	if (gf(a)!=gf(b)){
		fa[gf(a)]=gf(b);
	}
}
int bd;
int main() {
	// freopen("b.in","r",stdin);
	cin>>n>>k;
	for(int i=1;i<=n;i++)scanf("%d",&p[i]);
	for(int i=1;i<k;i++)s.insert(p[i]);
	int dz=1;
	for(int i=1;i+k-1<=n;i++){
		s.insert(p[i+k-1]);
		if (*s.begin() == p[i]) {
			ismin[i] = 1;
		}
		if (ismin[i - 1] && *s.rbegin() == p[i + k - 1]) {
			merge(i - 1, i);
		}
		dz=max(dz,i);
		while(dz<=n && p[dz+1]>p[dz])dz++;
		if(dz>=i+k-1){
			if(!bd)bd=i;
			else merge(bd,i);
		}
		s.erase(p[i]);
	}
	int ans=0;for(int i=1;i+k-1<=n;i++)if(gf(i)==i)ans++;
	cout<<ans<<endl;
}