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: Accepted)


int a[100009];
int s[100009];
int
main()
{
  int i;
  int ai;
  int n,m;
  scanf("%d %d", &n, &m);
  for (i=0; i<m; i++) {
    scanf("%d", &a[i]);
  }

  ai = 0;
  s[0] = 1;
  if (a[ai] == 1) {
    s[1] = 0;
    ai ++;
  } else {
    s[1] = 1;
  }
  for (i=1; i<n; i++) {
    if (a[ai] == i) {
      s[i] = 0;
      s[i+1] = s[i-1];
      ai++;
    } else {
      s[i+1] = (s[i-1] + s[i]) % 1000000007;
    }
  }

  /*
  for (i=0; i<n+1; i++) {
    printf("%d\n", s[i]);
  }
  */
  printf("%d\n", s[n]);
  
}