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>
#define int long long

using namespace std;

signed main(){
    int N;
    int K;
    cin>>N>>K;
    int arr[N];
    for(int i=0;i<N;i++){
        cin>>arr[i];
    }
    int length = 1;
    vector<int> increasing_sub_array_lengths;
    map<int,int> freq_increasing_sub_array;
    int flag = 0;
    for(int i=0;i<N-1;i++){
        if(arr[i+1] >= arr[i] && flag == 0){
            flag = 1;
            // cout<<"Starting at"<<arr[i]<<endl;
            length++;
        }
        else if(flag==1 && i==N-1){
            flag = 0;
            // cout<<"Ending at"<<arr[i]<<endl;
            increasing_sub_array_lengths.push_back(length);
            length = 1;            
        }
        else if(flag == 1 && arr[i+1] >= arr[i]){
            length++;
            // cout<<"Continuing at"<<arr[i]<<endl;
        }
        else if(flag == 1 && arr[i+1] < arr[i]){
            flag = 0;
            // cout<<"Ending at"<<arr[i]<<endl;
            increasing_sub_array_lengths.push_back(length);
            length = 1;
        }

    }
    if(flag==1 && arr[N-1]>=arr[N-2]){
    	increasing_sub_array_lengths.push_back(length);
    }
    // for(int i=0;i<increasing_sub_array_lengths.size();i++){
    //     cout<<increasing_sub_array_lengths[i]<<" ";
    // }
    int ans = 0;
    ans = N-K;
    for(int i=0;i<increasing_sub_array_lengths.size();i++){
    	if(increasing_sub_array_lengths[i]>K){
    		ans = ans - (increasing_sub_array_lengths[i]-K);
    	}
    }
    cout<<ans;
}