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<vector>
#include<iostream>
#include<set>
#include<algorithm>
using namespace std;

int main(){
  int n,k; cin >> n >> k;
  vector<int> p(n);
  for(int i=0;i<n;i++) cin >> p[i];
  set<int> s;
  vector<bool> c(n, true);
  int cs=0,cg=0;
  for(int i=0;i<k-2;i++){
    if(p[i] < p[i+1]) cs++;
    else cg++;
  }
  for(int i=k-1;i<n;i++){
  	if(p[i-1] < p[i]) cs++;
    else cg++;
    if(cg==0) c[i]=false;
	
    if(p[i-1-k+2] < p[i-k+2]) cs--;
    else cg--;
  }
  int ans=1;
  bool nor = false;
  if(!c[k-1]){
    nor = true;
    ans = 0;
  }
  for(int i=0;i<k;i++) s.insert(p[i]);
  for(int i=k;i<n;i++){
  	s.insert(p[i]);
    auto lit = s.begin();
    auto rit = s.end(); rit--;
  	if(*lit == p[i-k] && *rit == p[i]) ;
    else if(c[i]) ans++;
    else nor = true;
    
    s.erase(p[i-k]);
  }
  if(nor) ans++;
  cout << ans << endl;
  return 0;
}