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_base::sync_with_stdio(false); cin.tie(0);
  int N, K; cin >> N >> K;
  vector<int> A(N), B;
  int ans = 1;
  set<int> S;
  for (int i = 0; i < N; ++i) {
    cin >> A[i]; 
  }
  for (int i = 0; i < K; ++i) {
    S.insert(A[i]);
  }
  for (int i = K; i < N; ++i) {
    bool id = true;
    id &= A[i - K] == *S.begin();
    S.erase(A[i - K]);
    S.insert(A[i]);
    id &= A[i] == *S.rbegin();
    if (not id) ++ans;
  }
  vector<int> ok(N, 1);
  for (int i = 1; i < N; ++i) {
    if (A[i - 1] < A[i]) ok[i] = ok[i - 1] + 1;
    else ok[i] = 1;
  }
  bool first = 0;
  for (int i = K - 1; i < N; ++i) {
    if (first and ok[i] == K) --ans;
    first |= ok[i] >= K;
  }
  cout << ans << '\n';
}
