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>
#define abs(a) ((a) > 0 ? (a) : -(a))

int
main(int argc, char *argv[])
{
  int n, m, a[100000], d[100000];
  scanf("%d %d", &n, &m);
  for(int i = 0; i < m; i++) scanf("%d", &a[i]);

  d[0] = a[0] - 1;
  for(int i = 1; i < m; i++){
    d[i] = a[i] - a[i - 1] - 2;
  }
  d[m] = n - a[m - 1] - 1;
 
  long long x[100005]; x[0] = 1; x[1] = 2;
  for(int i = 2; i <= n; i++) {
    x[i] = x[i - 1] + x[i - 2];
    x[i] = x[i] % 1000000007;
  }

  long long ans = 1;

  for(int i = 0; i <= m; i++){
    if(d[i] < 0){ans = 0; break;}
    else if(d[i] == 0) continue;
    else {
      ans = (ans * x[d[i] - 1]);
      ans = ans % 1000000007;
    }
  }

  printf("%lld\n", ans);

  return 0;
}