#include <bits/stdc++.h>
typedef long long lint;
using namespace std;
int main(){
  int N,K;
  cin>>N>>K;
  int P[N];
  for(int i=0;i<N;i++)cin>>P[i];
  set<int> s;
   for(int i=0;i<K;i++)s.insert(P[i]);
  int ans=1;
  for(int i=0;i<N-K;i++) {
    if(P[i]==*s.begin() and P[i+K]>*s.rbegin())ans--;  
    // cerr<<*s.begin()<<" "<<*s.rbegin()<<endl;
    ans++;
    s.erase(P[i]);
    s.insert(P[i+K]);
  }    //相交区间   单调队列 
  int a=0,b=1;
 for(int i=1;i<N;i++)
  {
    if(P[i-1]<P[i])b++;
    else b=1;
    if(b==K)a++;
  }  
  ans-=max(a-1,0);    //不相交区间 
  cout<<ans<<endl;
  return 0;
}

### Prompt

Generate a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
typedef long long lint;
using namespace std;
int main(){
  int N,K;
  cin>>N>>K;
  int P[N];
  for(int i=0;i<N;i++)cin>>P[i];
  set<int> s;
   for(int i=0;i<K;i++)s.insert(P[i]);
  int ans=1;
  for(int i=0;i<N-K;i++) {
    if(P[i]==*s.begin() and P[i+K]>*s.rbegin())ans--;  
    // cerr<<*s.begin()<<" "<<*s.rbegin()<<endl;
    ans++;
    s.erase(P[i]);
    s.insert(P[i+K]);
  }    //相交区间   单调队列 
  int a=0,b=1;
 for(int i=1;i<N;i++)
  {
    if(P[i-1]<P[i])b++;
    else b=1;
    if(b==K)a++;
  }  
  ans-=max(a-1,0);    //不相交区间 
  cout<<ans<<endl;
  return 0;
}

```