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;
int main(){
  int N, K;
  cin >> N >> K;
  vector<int> vec(N);
  for(int i=0; i<N; i++){
    cin >> vec[i];
  }
  vector<int> bigger(N-1,0);
  for(int i=0; i<N-1; i++){
    if(vec[i] < vec[i+1]){
      bigger[i] = 1;
    }
  }
  
  set<int> S;
  int ans = 1;
  int cnt = 0;
  for(int i=0; i<K; i++){
    S.insert(vec[i]);
    if(i != 0){
      cnt += bigger[i-1];
    }
  }
  
  bool arrange = false;
  
  if(cnt == K-1){
    arrange = true;
  }
  
  

  
  for(int i=K; i<N; i++){
    auto itr_min = S.begin();
    S.insert(vec[i]);
    auto itr_max = S.rbegin();
    cnt -= bigger[i-K];
    cnt += bigger[i-1];
    if(*itr_min != vec[i-K] || *itr_max != vec[i]){
      if(cnt != K-1){
        ans++;
      }
      else if(!arrange){
        ans++;
        arrange = true;
      }
    }
    S.erase(vec[i-K]);
  }
  cout << ans << endl;
}