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 (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC optimize("Ofast")
#include <stdio.h>

main()
{
    int n,k; scanf("%d%d",&n,&k);
    int ary[1<<18];
    int que[1<<18];
    int que2[1<<18];
    int *beg=que,*end=que;
    int *beg2=que2,*end2=que2;
    int sorted=0;
    int ans=0;

    for(int i=0; i<n; ++i)
    {
        scanf("%d",ary+i);
        int flag=0;

        if(i>=k)
        {
            if(*beg==i-k) ++beg;
            else flag=1;

            if(*beg2==i-k) ++beg2;
        }

        while(beg!=end && ary[*(end-1)]>ary[i])
        {
            --end;
        }
        *end++=i;

        while(beg2!=end2 && ary[*(end2-1)]<ary[i]) --end2;
        if(beg2!=end2) flag=1;
        *end2++=i;

        if(i==k-1) flag=1;
        else if(i<k-1) flag=0;

        if(end-beg==k)
        {
            flag=0;
            sorted=1;
        }

        if(flag)
        {
            ans++;
            // printf("%d\n",i);
        }
    }

    ans+=sorted;
    printf("%d\n",ans);
}