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
cmp(const void *a, const void *b)
{
    return *(int *)(a)-*(int *)(b);
}

int
main(int argc, char *argv[])
{
    int N, K;
    int P[200000];
    int Q[200000];
    int R[200000];
    int i, j;
    int num=0;

    scanf("%d%d", &N, &K);
    for (i=0; i<N; i++) {
        scanf("%d", P+i);
    }
    if (N==K) {
        printf("1\n");
        return 0;
    }
    bcopy(P, R, N*sizeof(int));
    for (i=0; i<N-K+1; i++) {
        bcopy(P, Q, N*sizeof(int));
        qsort(Q+i, K, sizeof(int), cmp);
#if 0
        printf("%d: ", i);
        for (j=0; j<N; j++)
            printf("%d ", Q[j]);
        printf("\n");
#endif
        if (i==0){
                bcopy(Q, R, N*sizeof(int));
                num++;
                continue;
        }
        for (j=i; j<i+K; j++) {
            if (Q[j]!=R[j]) {
#if 0
                printf("%d, %d: %d %d\n", i, j, Q[j], R[j]);
#endif                
                bcopy(Q, R, N*sizeof(int));
                break;
            }
        }

        if (j!=i+K) {
#if 0
            printf("++ %d\n", i);
#endif
            num++;
        }
    }
    printf("%d\n", num);
    return 0;
}