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<iostream>
#include<algorithm>
#include<vector>
#include<set>
using namespace std;
typedef long long ll;
#define chmin(a,b) a=min(a,b)
#define chmax(a,b) a=max(a,b)
#define mod 1000000007
ll n,k,p[200010];
int main(){
  cin>>n>>k;
  for(int i=0;i<n;i++){
      cin>>p[i];
  }
  ll cnt=0;
  for(int i=0;i<k-1;i++){
      cnt+=(p[i]>p[i+1]);
  }
  set<int> st;
  for(int i=0;i<=k;i++){
      st.insert(p[i]);
  }
  ll ans=n-k+1;
  bool th=0;
  for(int i=0;i<n-k;i++){
      if(cnt==0){
	  ans--;
	  th=1; //cout<<i<<"*"<<endl;
      }
      else{
	  auto it=st.end(); it--;
	  if(p[i]==*st.begin()&&p[i+k]==*it){
	      ans--;
	  }
      }
      st.erase(p[i]);
      st.insert(p[i+k+1]);
      cnt-=(p[i]>p[i+1]);
      cnt+=(p[i+k-1]>p[i+k]);
  }
  if(cnt==0)ans--,th=1;
  ans+=th;
  cout<<ans<<endl;
  return 0;
}
