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<set>
using namespace std;
#define df 0

int main(){
  if(df) printf("*debug mode*\n");
  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]);
  }
  if(df){
    printf("s: ");
    for(int x: s){ printf("%d ",x); }
    printf("\n");
  }
  int a=0,cnt=1;
  for(int i=k;i<n;i++,a++){
    if(df)printf("%d %d\n",*(s.begin()),*(s.rbegin()));
    if(*(s.begin())==p[a] && *(s.rbegin())<p[i]){cnt--;}
    cnt++;
    s.erase(p[a]);
    s.insert(p[i]);
  }
  int dup=0,inc=1;
  for(int i=1;i<n;i++){
    if(p[i-1]<p[i]) inc++;
    else inc=1;
    if(df)printf("%d ",inc);
    if(inc==k)dup++;
  }
  dup--;
  if(df)printf("\n%d\n",dup);
  if(dup>0)cnt-=dup;
  cout << cnt;
}

/// confirm df==0 ///
