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<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
#include<set>

using namespace std;

#define MOD 1000000007

int main(){
  long long ans = 0;
  int n, k, t;
  cin >> n >> k;
  vector<pair<int, int>> p(n+1);
  set<pair<int, int>> range;
  p[0] = {-1, 0};
  range.insert({-1, 0});
  bool flag = false;
  vector<int> great(n+1, 0);
  for(int i = 1; i <= n; i++){
    cin >> t;
    p[i] = {t, i};
    range.insert({t, i});
    if(p[i].first > p[i-1].first){
      great[i] = great[i-1]+1;
    }
    if(i > k-1){
      auto it = range.find(p[i-k]);
      if(i == k){
        if(great[i] - great[i-(k-1)] == k-1) flag = true;
        ans++;
      }
      else if(great[i] - great[i-(k-1)] == k-1){
        if(!flag) ans++, flag = true;
      }
      else if(it != range.begin() || range.rbegin()->second != i){
        ans++;
      }
      range.erase(it);
    }
  }
  cout << ans << endl;
}