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 sz[200001], par[200001];
int find(int idx){
    while(idx!=par[idx]){
        idx=par[idx];
    }
    return idx;
}
void join(int x, int y){
    if(x==y) return;
    if(sz[x]<sz[y]){
        swap(x,y);
    }
    sz[x]+=sz[y];
    sz[y]=0;
    par[y]=x; 
}
int main() {
	// your code goes her
	int n, k, flag=0, cnt[200001],p[200001],ans=0; 
	cin>>n>>k;
  cnt[1]=0;
	for(int i=1; i<=n; i++){
	    cin>>p[i]; 
	    if(i<=n-k+1) sz[i]=1,par[i]=i; 
	    if(i>1&&p[i]>p[i-1])cnt[i]=cnt[i-1];
	    cnt[i]++;
	    if(cnt[i]>=k) flag=1;
	}
	set<int> s; 
	for(int i=1; i <=k; i++){
	    s.insert(p[i]);
	}
	for(int i=k+1;i<=n;i++){
	    s.erase(p[i-k]);
	    if(p[i-k]< *s.begin() && *s.rbegin() < p[i]) join(find(i-k),find(i-k+1)); 
	    s.insert(p[i]);
	}
	for(int i=1;i<=n-k+1;i++) ans+=(sz[i]!=0 && cnt[i+k-1]<k);
	cout<<ans+flag<<endl;
	return 0;
}