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 <set>

int n, k;
int P[200000];


int main(){
  int i, r;
  std::set<int> s;
  scanf("%d%d", &n, &k);
  for(i=0;i<n;i++) scanf("%d", P+i);
  int sort = 0;
  for(i=0;i<k;i++){
    s.insert(P[i]);
    sort += (i!=0&&P[i] > P[i-1]);
  }
  r = 1;
  int m = sort == k-1;
  for(i=k;i<n;i++){
    if(P[i-k] == *s.begin() && P[i] > *s.rbegin()) ;
    else r++;
    int nsort = sort - (P[i-k+1] > P[i-k]) + (P[i] > P[i-1]);
    m += sort != k-1 && nsort == k-1;
    sort = nsort;

    s.erase(P[i-k]);
    s.insert(P[i]);
  }
  printf("%d\n", r - (m?m-1:0));

  return 0;
}
