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: Accepted)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn = 200005;
struct node{
    int value;
    int index;
    bool operator<(const node & t)const{
        return value < t.value;
    }
};
struct node2{
    int value;
    int index;
    bool operator<(const node2 & t)const{
        return value > t.value;
    }
};
priority_queue<node>q1;
priority_queue<node2>q2;
int a[maxn];
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++)
        scanf("%d",&a[i]);
    int lianxu=1;
    int lianxunum=0;
    int flag=0;
    for(int i=1;i<n;i++){
        if(a[i]>=a[i-1]){
            lianxu++;
            if(lianxu==k)
                lianxunum++;
        }
        else
            lianxu=1;
    }
    for(int i=0;i<k;i++){
        q1.push({a[i],i});
        q2.push({a[i],i});
    }
    int ans = 1;
    //cout << q1.top().value << q2.top().value <<endl;
    for(int i=k;i<n;i++){
        ans++;
        while(q1.top().index<i-k){
            q1.pop();
        }
        while(q2.top().index<i-k){
            q2.pop();
        }
        if(a[i]>=q1.top().value){
            if(q2.top().value==a[i-k]){
                ans--;
            }
        }
        q1.push({a[i],i});
        q2.push({a[i],i});
    }
    lianxunum=max(0,lianxunum-1);
    printf("%d\n",ans-lianxunum);
    return 0;
}