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 <iostream>
#include <vector>
#include <algorithm>

using namespace std;

void copy(vector<int> &dst, vector<int> &src, int from, int size)
{
    for (int i = 0; i < size; ++i)
    {
        dst[i] = src[i + from];
    }
}

int main()
{
    int N, K;
    cin >> N >> K;

    vector<int> P(N);
    for (int &n : P) cin >> n;

    vector<int> partial(K);
    copy(partial, P, 0, K);
    sort(partial.begin(), partial.end());
    vector<int> partialPre = partial;

    int count = 1;
    for (int i = 0; i < N - K; ++i)
    {
        copy(partial, P, i, K);
        sort(partial.begin(), partial.end());

        if (
            (partial[0] != P[i] && partialPre[1] != partial[0]) || partial[K - 1] != P[i + K - 1]
        )
        {
            ++count;
        }

        partialPre = partial;
    }

    cout << count << endl;

    return 0;
}