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 <cstdio>
#include <vector>
#include <deque>
using namespace std;


vector<int> slidemin(int n,int m,vector<int> &A)
{
  deque<int> Q;
  vector<int> B(n,0);
  for(int i=0;i<n;i++){
    while(!Q.empty()&&A[Q.back()]>=A[i]){
      Q.pop_back();
    }
    Q.push_back(i);
    if(i>=m&&Q.front()==i-m){
      Q.pop_front();
    }
    if(i>=m-1){
      B[i]=A[Q.front()];
    }
  }
  return B;
}

int main()
{
  int n,k;
  scanf("%d%d",&n,&k);
  vector<int> a(n),ma(n);
  for(int i=0;i<n;i++){
    scanf("%d",&a[i]);
    ma[i]=-a[i];
  }
  vector<int> b,mb,L(n);
  b=slidemin(n,k-1,a);
  mb=slidemin(n,k-1,ma);
  L[0]=1;
  for(int i=1;i<n;i++){
    L[i]=(a[i-1]<=a[i]?L[i-1]+1:1);
  }
  int ans=0;
  bool F=false;
  for(int i=k-1;i<n;i++){
    if(L[i]>=k){
      if(!F){
	F=true;
	ans++;
      }
    }
    else if(!(i>=k&&a[i-k]<=b[i-1]&&-mb[i-1]<=a[i])){
      ans++;
    }
  }
  printf("%d\n",ans);
  return 0;
}
