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: Wrong Answer)

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

int
min(int *a, int num)
{
    int i;
    int ret;
    
    ret = a[0];
    for (i=1; i<num; i++) {
        if (a[i]<ret)
            ret = a[i];
    }
    return ret;
}

int
max(int *a, int num)
{
    int i;
    int ret;
    
    ret = a[0];
    for (i=1; i<num; i++) {
                if (a[i]>ret)
            ret = a[i];
    }
    return ret;
}

int
sorted(int *a, int num)
{
    int i;

    for (i=0; i<num-1; i++)
        if (a[i]>a[i+1])
            return 0;
    return 1;
}

int
main(int argc, char *argv[])
{
    int N, K;
    int P[200002];
    int min_p, max_p;
    int i, j;
    int num=0;
    int num_sort=0;

    scanf("%d%d", &N, &K);
    for (i=0; i<N; i++) {
        scanf("%d", P+i);
    }
    if (N==K) {
        printf("1\n");
        return 0;
    }

    max_p = max(P+1, K);
    for (i=0; i<N-K+1; i++) {
#if 0
        for (j=0; j<K; j++) {
            printf("%d ", *(P+i+j));
        }
        printf("\n");
#endif
        if (sorted(P+i, K)==1) {
            num_sort++;
        }
        else {
            min_p = min(P+i, K);
#if 0
            printf("min_p %d, *(P+%d) %d, max_p %d, *(P+%d+K(%d)) %d ", min_p, i, *(P+i), max_p, i, K, *(P+i+K));
#endif
            if (min_p!=*(P+i)) {
#if 0                
                printf("*");
#endif                
                num++;
                continue;
            }
#if 0            
            printf("\n");
#endif
            max_p = max(P+i, K);
            if (max_p>P[i+K]) {
                num++;
                continue;
            }
        }
    }
#if 0
    printf("num %d s %d\n", num, num_sort);
#endif    
    num = num + (num_sort>0?1:0);

    printf("%d\n", num);
    return 0;
}