Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

5 3
0 2 1 4 3

Sample Output 1

2

Two permutations can be produced as P after the operation: (0,1,2,4,3) and (0,2,1,3,4).

Sample Input 2

4 4
0 1 2 3

Sample Output 2

1

Sample Input 3

10 4
2 0 1 3 7 5 4 6 8 9

Sample Output 3

6

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll=long long;
#define MAX 500009
#define MOD 1000000007
#define INF 1000000000000000000

int main(){
  int N,K;
  cin>>N>>K;
  vector<int> P(N);
  for(int i=0;i<N;i++){
    cin>>P.at(i);
  }

  int ans=1;
  set<int> s;
  for(int i=0;i<K;i++){
    s.insert(P.at(i));
  }
  for(int i=K;i<N;i++){
    int m=*s.begin();
    s.erase(m);
    s.insert(P.at(i));
    int M=*s.rbegin();
    if(m==P.at(i-K)&&M==P.at(i)){
      continue;
    }else{
      ans++;
    }
  }

  int same=0;
  int count=1;
  for(int i=1;i<N;i++){
    if(P.at(i-1)<P.at(i)){
      count++;
    }else{
      if(count-K>=0){
        same++;
      }
      count=1;
    }
  }
  if(count-K>=0){
    same++;
  }
  ans-=max(same-1,0);

  cout<<ans<<endl;
}