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)

#pragma GCC optimize ("O3")
#include <iostream>
#include <iomanip>
#include <istream>
#include <ostream>
#include <sstream>
#include <iterator>
#include <vector>
#include <algorithm>
#include <queue>
#include <deque>
#include <list>
#include <stack>
#include <map>
#include <unordered_map>
#include <set>
#include <bitset>
#include <utility>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <ctime>
#include <cctype>
#include <cstdlib>
#define IINF 100000000
#define INF 300000000000000000
#define MOD 1000000007
#define mod 1000000007
#define INT_MAX_ 2147483647
#define EPS (1e-10)
#define REP(i, a, n) for (ll i = a; i < (ll)(n); i++)
#define REPE(i, a, n) for (ll i = a; i <= (ll)(n); i++)
#define rep(i,n)for (ll i = 0; i < (ll)(n); i++)
#define Endl endl
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define eb emplace_back
#define mmax(x,y)(x>y?x:y)
#define mmin(x,y)(x<y?x:y)
#define chmax(x,y) x=mmax(x,y)
#define chmin(x,y) x=mmin(x,y)
#define all(x) (x).begin(),(x).end()
#define siz(x) (ll)(x).size()
#define PI acos(-1.0)
#define me memset
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
typedef pair<int,int>Pin;
typedef pair<ll,ll>Pll;
template<class T> using V=vector<T>;
template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T> >;
long long GCD(long long a, long long b) {return b?GCD(b,a%b):a;}
long long LCM(long long a, long long b) {return a/GCD(a,b)*b;}
int dx[4]={-1,0,1,0};
int dy[4]={0,-1,0,1};
int ddx[8]={-1,0,1,0,1,1,-1,-1};
int ddy[8]={0,-1,0,1,1,-1,1,-1};
ll cmp(pair<ll,Pll> a,pair<ll,Pll> b){
        if(a.fi!=b.fi)
        return a.fi<b.fi;
        else
        return a.se.fi>b.se.fi;
}
//----------------------------------------------------------------------

//----------------------------------------------------------------------
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    //-------------------------------  
    //ll begin_t=clock();
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    //-------------------------------
    ll n,k;cin>>n>>k;
    V<ll>p(n);
    for(ll i=0;i<n;i++){
        cin>>p[i];
    }
    map<Pll,ll>M;

    ll l=0,r=0;
    ll ccc=0;
    for(ll i=1;i<n;i++){
        if(p[i]>p[i-1]){
            r=i;
        }
        else{
            for(ll j=l;j+k-1<=r;j++){
                M[{j,j+k-1}]++;
                ccc++;
            }
            l=i;
        }
    }
    for(ll j=l;j+k-1<=r;j++){
        M[{j,j+k-1}]++;
        ccc++;
    }
 
    priority_queue<ll,vector<ll>,greater<ll>>Q1;
    priority_queue<ll>Q2;
    ll minm=INF,maxm=-1;
    for(ll i=0;i<k;i++){
        chmin(minm,p[i]);
        chmax(maxm,p[i]);
        Q1.push(p[i]);
        Q2.push(p[i]);
    }
    l=p[0];
    ll ans=1;
    bool abc=0;
    for(ll i=k;i<n;i++){
        r=p[i];
        ll tmp1=Q1.top();
        if(tmp1==minm){
            Q1.pop();
            chmax(minm,Q1.top());
        }
        ll tmp2=Q2.top();
        if(tmp2==maxm){
            chmin(maxm,Q2.top());
        }
        if(r<maxm||l>minm){
            ans++;
        }
        Q1.push(p[i]);
        Q2.push(p[i]);
        chmin(minm,p[i]);
        chmax(maxm,p[i]);
        l=p[i-k+1];
    }
    //ans=ans-mmax((ccc-1),0);
    cout<<ans<<endl;
    //-------------------------------  
    //fclose(stdin);
    //fclose(stdout);
    //ll end_t=clock();cout<<"time="<<end_t-begin_t<<"ms"<<endl;
    //-------------------------------
    return 0;
}
//----------------------------------------------------------------------