Problem:
Score : 300 points

Problem Statement

There is a staircase with N steps. Takahashi is now standing at the foot of the stairs, that is, on the 0-th step.
He can climb up one or two steps at a time.

However, the treads of the a_1-th, a_2-th, a_3-th, \ldots, a_M-th steps are broken, so it is dangerous to set foot on those steps.

How many are there to climb up to the top step, that is, the N-th step, without setting foot on the broken steps?
Find the count modulo 1\ 000\ 000\ 007.

Constraints

1 \leq N \leq 10^5

0 \leq M \leq N-1

1 \leq a_1 < a_2 < ... < a_M \leq N-1

Input

Input is given from Standard Input in the following format:

N M
a_1
a_2
.
.
.
a_M

Output

Print the number of ways to climb up the stairs under the condition, modulo 1\ 000\ 000\ 007.

Sample Input 1

6 1
3

Sample Output 1

4

There are four ways to climb up the stairs, as follows:

0 \to 1 \to 2 \to 4 \to 5 \to 6

0 \to 1 \to 2 \to 4 \to 6

0 \to 2 \to 4 \to 5 \to 6

0 \to 2 \to 4 \to 6

Sample Input 2

10 2
4
5

Sample Output 2

0

There may be no way to climb up the stairs without setting foot on the broken steps.

Sample Input 3

100 5
1
23
45
67
89

Sample Output 3

608200469

Be sure to print the count modulo 1\ 000\ 000\ 007.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
int acs(const void *a, const void *b) { return *(int *)a - *(int *)b; }                            /* 1,2,3,4.. */
int des(const void *a, const void *b) { return *(int *)b - *(int *)a; }                            /* 8,7,6,5.. */
int cmp_char(const void *a, const void *b) { return *(char *)a - *(char *)b; }                     /* a,b,c,d.. */
int cmp_str(const void *a, const void *b) { return strcmp(*(const char **)a, *(const char **)b); } /* aaa,aab.. */
#define min(a, b) (a < b ? a : b)
#define max(a, b) (a > b ? a : b)
#define mod 1000000007

int main(void)
{
  int n, m, broken[100001] = {0};
  scanf("%d %d", &n, &m);

  int bridge;
  int i;
  for (i = 0; i < m; i++)
  {
    scanf("%d", &bridge);
    broken[bridge] = 1;
  }

  int dp[100001];
  dp[0] = 1;
  for (i = 0; i < n; i++)
  {
    if (broken[i] == 0)
    {
      if (broken[i + 1] == 0)
        dp[i + 1] = (dp[i + 1] + dp[i]) % mod;
      if (i + 2 <= n && broken[i + 2] == 0)
        dp[i + 2] = (dp[i + 2] + dp[i]) % mod;
    }
  }

  printf("%d\n", dp[n] % mod);

  return 0;
}