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 <algorithm>
#include <cassert>
#include <cmath>
#include <iostream>
#include <map>
#include <queue>
#include <vector>

static const int IINF = 1 << 30;
static const long long LINF = 1LL << 60;
static const long long MOD = 1.0e+9 + 7;

template <typename T> std::vector<T> vectors(std::size_t n, T val) {
  return std::vector<T>(n, val);
}

template <typename T, typename... Args>
auto vectors(std::size_t n, Args... args) {
  return std::vector<decltype(vectors<T>(args...))>(n, vectors<T>(args...));
}

template <class T> inline bool chmin(T &a, const T &b) {
  return (a > b) ? a = b, true : false;
}

template <class T> inline bool chmax(T &a, const T &b) {
  return (a < b) ? a = b, true : false;
}

template <class T> inline void chadd(T &a, const T &b) {
  a += b, a %= MOD;
  // TODO minus case
}

template <class T>
std::ostream &operator<<(std::ostream &s, const std::vector<T> &v) {
  if (v.empty())
    return s;
  s << *v.begin();
  for (auto iter = v.begin() + 1; iter != v.end(); ++iter)
    if (std::is_fundamental<T>::value)
      s << " " << *iter;
    else
      s << std::endl << *iter;
  return s;
}

int main() {
  int N, K;
  std::cin >> N >> K;
  std::vector<int> P(N);
  for (int i = 0; i < N; ++i)
    std::cin >> P[i];

  int upCount = 0;
  for (int i = 0; i < N - K + 1; ++i) {
    bool up = true;
    for (int k = 0; k < K - 1; ++k) {
      if (P[i + k] > P[i + k + 1]) {
        up = false;
        break;
      }
    }
    if (up)
      upCount += 1;
  }
  // std::cout << upCount << std::endl;


  int res = N - K + 1;
  if (upCount >= 2)
    res -= (upCount - 1);

  std::cout << res << std::endl;

  return 0;
}