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++ (GCC 5.4.1) (judge verdict: Wrong Answer)

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

int main(){
    int N,K;
    cin >> N >> K;
    int P[N];
    for(int i=0;i<N;++i){
        cin >> P[i];
    }
    vector<int>q;
    q.insert(q.begin(),P[0]);
    for(int i=1;i<K;++i){
        q.insert(lower_bound(q.begin(),q.end(),P[i]),P[i]);
    }
   
    int c=1;
    if(N==K){
        cout<<1;
        return 0;
    }
    for(int i=K;i<N;++i){
        if(P[i]>q[K-1] && P[i-K]==q[0]){
            c+=1;
        }
        q.erase(lower_bound(q.begin(),q.end(),P[i-K]));
        q.insert(lower_bound(q.begin(),q.end(),P[i]),P[i]);
    }
    cout << c;
    return 0;
}