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: Runtime Error)

#include<stdio.h>
#include<stdlib.h>


int main(void){

int sort=0;
int min,max;
int Min,Max;
int min_bef=-1;
int Def=0;
int x=0;
int k_n;
int *k_p;
int n;
int i,j,k;
int *p;
int *p_s;

(void)scanf("%d %d",&n,&k_n);
p = malloc(sizeof(*p)*n);
p_s = p;
  
        for(i=0;i<n;i++){
        (void)scanf("%d",p);
        p++;
        }
k_p = p_s;

        for(i=0;i<n-k_n+1;i++){
        min = k_p[0];
        max = k_p[k_n-1];
                for(j=0;j<k_n-1;j++){
                        if(k_p[j]<k_p[j+1]){
                        Min=k_p[i];
                        Max=k_p[i+1];
                        sort++;
                        }else{

                        Min=k_p[i+1];
                        Max=k_p[i];
                        }
                if(min>Min)
                min=-1;

                if(max<Max)
                max=-1;
                }

/*--------------------------------------------*/
                if(sort == k_n){
                x++;
                        if(Def==0){
                        Def=-1;
                        }else {
                        x--;
                        }
                }else{
                x++;
                        if((min_bef!=-1) && (max != -1)){
                        x--;
                        }
                }
        sort=0;
        min_bef=min;
        k_p=&(p_s[i+1]);
        }
printf("%d",x);

return 0;

}