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;
const int maxn = 200011;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  vector<int> p(n), pre(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &p[i]);
  }
  for (int i = 1; i < n; ++i) {
    if (p[i] < p[i - 1]) {
      pre[i] = i;
    } else {
      pre[i] = pre[i - 1];
    }
  }
  // for (int i = 0; i < n; ++i) {
  //   printf("%d%c", pre[i], " \n"[i == n - 1]);
  // }
  set<int> st;
  for (int i = 0; i < k; ++i) {
    st.insert(p[i]);
  }
  int ans = 1;
  int num = (pre[k - 1] == 0);
  for (int i = k; i < n; ++i) {
    int mn = *st.begin(), mx = *st.rbegin();
    if (p[i - k] != mn || p[i] < mx) {
      ++ans;
      num += (pre[i] <= i - k + 1);
    }
    st.erase(p[i - k]);
    st.insert(p[i]);
  }
  if (num > 1) {
    ans -= (num - 1);
  }
  printf("%d\n", ans);
  return 0;
}
