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() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);

  int n, k;
  cin >> n >> k;
  vector<int> p(n);
  for (int i = 0; i < n; ++i) {
    cin >> p[i];
  }
  int ans = 1;
  set<int> cur;
  for (int i = 0; i < k; ++i) {
    cur.insert(p[i]);
  }
  multiset<int> diff;
  for (int i = 1; i < k; ++i) {
    diff.insert(p[i] - p[i - 1]);
  }
  int sorted = *diff.begin() > 0;
  for (int i = k; i < n; ++i) {
    bool ok = p[i - k] > *cur.begin() or p[i] < *cur.rbegin();
    if (ok) {
      ++ans;
    }
    cur.erase(p[i - k]);
    cur.insert(p[i]);
    bool ok2 = ok or *diff.begin() < 0;
    diff.erase(diff.find(p[i - k + 1] - p[i - k]));
    diff.insert(p[i] - p[i - 1]);
    if (ok2) {
      sorted += *diff.begin() > 0;
    }
  }
  cout << ans - max(0, sorted - 1);

  return 0;
}