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 <math.h>
#include <assert.h>

#include <set>
#include <vector>
#include <numeric>
#include <iostream>
#include <algorithm>
#include <functional>

template <typename T1, typename T2>
std::istream& operator>>(std::istream& is, std::pair<T1, T2>& v) {
    return is >> v.first >> v.second;
}

template <typename T>
std::istream& operator>>(std::istream& is, std::vector<T>& v) {
    for (T& i: v) is >> i;
    return is;
}

template <typename T1, typename T2>
std::ostream& operator<<(std::ostream& os, const std::pair<T1, T2>& v) {
    return os << v.first << " " << v.second;
}

template <typename T>
std::ostream& operator<<(std::ostream& os, const std::vector<T>& v) {
    int n = (int) v.size();
    for (int i = 0; i < n; ++i) os << v[i] << (i == n - 1 ? "" : " ");
    return os;
}

template <typename X>
void in(X& x) {
    std::cin >> x;
}

template <typename X, typename Y>
void in(X& x, Y& y) {
    in(x); in(y);
}

template <typename X, typename Y, typename Z>
void in(X& x, Y& y, Z& z) {
    in(x); in(y, z);
}

template <typename X, typename Y, typename Z, typename W>
void in(X& x, Y& y, Z& z, W& w) {
    in(x); in(y, z, w);
}

template <typename T>
void out(const T H) {
    std::cout << (H);
}

template <typename T1, typename... T2>
void out(const T1 H, const T2... T) {
    std::cout << (H) << " "; out(T...);
}

template <typename T>
void eout(const char *s, const T H) {
    std::cerr << s << "=" << H << "\n";
}

template <typename T1, typename... T2>
void eout(const char *s, const T1 H, const T2... T) {
    while (*s != ',') std::cerr << *s++;
    std::cerr << "=" << H << " | "; eout(++s, T...);
}

#define nl                std::cout << "\n"
#define err(args...)      eout(#args, args)

#define rep(i, n)         for (int i =  0 ; i < (n); ++i)
#define per(i, n)         for (int i = (n) - 1; i > -1; --i)
#define repr(i, l, r)     for (int i = (l); i < (r); ++i)

#define sz(v)             ((int) v.size())
#define all(v)            (v).begin(), (v).end()
#define lla(v)            (v).rbegin(), (v).rend()

#define no                std::cout << "NO\n"
#define yes               std::cout << "YES\n"

const int INF           = 2147483647;
const long long int LNF = 9223372036854775807LL;

using ll  = long long int;
using vi  = std::vector<int>;
using vb  = std::vector<bool>; //try not to use this
using ii  = std::pair<int, int>;
using iii = std::pair< ii, int>;
using vii = std::vector<ii>;

const int N  = 505050;
const int MD = 1000*1000*1000 + 7;

int solve();
int main() {
    std::ios_base::sync_with_stdio(0); std::cin.tie(0);

    //int t; in(t); repr(tc, 1, t + 1)
    solve();

    return 0;
}

int solve() {
    int n; in(n);
    int k; in(k);
    vi a(n); in(a);

    int i = 0, ans = 0;
    while (i < n) {
        int j = i + 1;
        while (j < n and a[j - 1] < a[j]) ++j;
        ans += std::max(j - i - k + 1, 0);
        i = j;
    }
    if (ans) --ans;
    ans = n - k + 1 - ans;
    if (k + 1 <= n) {
        std::set<int> s;
        for (i = 0; i < k + 1; ++i) {
            s.insert(a[i]);
        }
        for (i = 0; i + k + 1 < n; ++i) {
            if (*s.begin() == a[i] and *s.rbegin() == a[i + k])
                --ans;
            s.erase(a[i]);
            s.insert(a[i + k + 1]);
        }
    }
    out(ans);
    return 0;
}