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

Solution (in CPP):

Solution:
#include <stdio.h>
#include <set>
typedef long long ll;

int main(void) {
  ll i, j, k, n, ans = 0, cnt = 0;
  scanf("%lld%lld", &n, &k);
  bool flg = false, b[n];
  ll a[n];
  std::set<ll> s;
  std::set<ll>::iterator it;
  for(i = 0; i < n; ++i) {
    scanf("%lld", &a[i]);
    if(i < k) s.insert(a[i]);
    if(i) {
      if(a[i] > a[i - 1]) b[i] = false;
      else b[i] = true;
      if(i < k) cnt += b[i] ? 1 : 0;
    }
  }
  if(cnt) ++ans;
  else flg = true;
  for(i = k; i < n; ++i) {
    it = s.end(), it--;
    cnt += (b[i] ? 1 : 0) - (b[i - k + 1] ? 1 : 0);
    if((a[i] < *it || *s.begin() != a[i - k]) && cnt) ++ans;
    if(!cnt) flg = true;
    s.erase(s.find(a[i - k])), s.insert(a[i]);
  }
  printf("%lld", ans + (flg ? 1 : 0));
  return 0;
}