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)

#include <stdio.h>

int main() {
  int number_n, number_m, i, j = 0;
  scanf("%d %d", &number_n, &number_m);
  long long int stage[number_n+1];
  int broken_stage[number_m];
  for(i = 0; i < number_m; i++) scanf("%d", &broken_stage[i]);
  stage[0] = 1;
  stage[1] = (broken_stage[0] == 1)? 0 : 1;
  for(i = 2; i <= number_n; i++) {
    stage[i] = stage[i-1] + stage[i-2];
    for(; j < number_m; j++) {
      if(broken_stage[j] == i) {
        stage[i] = 0;
        break;
      } else if(broken_stage[j] > i) {
        break;
      }
    }
    if(stage[i] > 1000000007) stage[i] -= 1000000007;
  }
  printf("%lld\n", stage[number_n]);
  return 0;
}