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: Runtime Error)

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
long long int combination(int n, int r)
{
    if (r == 0 || r == n)
        return (1);
    else if (r == 1)
        return (n);

    long long int a=1, b=1;
    for(int i=0; i<r; i++){
      a *= n-i;
      b *= r-i;
    }
    return (a/b);
}



int main(){
  int N, M;
  scanf("%d %d", &N, &M);
  int a[M+1], i, j, S[N];
  long long int ans = 1;
  for(i=0; i<M; i++){
    scanf("%d", &a[i]);
  }

  int cnt=1;
  S[0] = a[0] - 2;
  for(i=0; i<M-1; i++){
    S[cnt] = 1;
    cnt++;
    S[cnt] = a[i+1] - a[i] - 2;
    if (S[cnt] < 0){
      printf("0");
      return 0;
    }
    cnt++;
  }
  S[cnt] = 1;
  cnt++;
  S[cnt] = N - a[i]-1;
  cnt++;

  long long int *K = (long long int*)calloc(N, sizeof(long long int));
  //long long int K[N];
  for(i=0; i<cnt; i++){
    if (S[i] != 0){
      for(j=0; j<(S[i]/2)+1; j++){
        K[i] += combination(S[i]-j, j);
      }
    }else{
      K[i] = 1;
    }
  }

  for(i=0; i<cnt; i++){
    ans *= K[i];
  }

  printf("%d", (int)(ans%(long long int)1000000007));
  
  return 0;
}