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>

int i;
int main(void){
  
  
  int now=0;
  long ans=1;
  int terget;
  int n,m;//top,breaking
  int breaker;
  int max=0;
  scanf("%d %d",&n, &m);
  int inter[m+1];
  for(i=0;i<m;i++){
    scanf("%d",&breaker);
    inter[i]=breaker - now-1;
    //printf("%d\n",inter[i]);
    if(inter[i]==-1){
        ans=0;

    }
    if(inter[i] >max){
      max=inter[i];
    }
    now=breaker+1;
  }
  inter[m]=n-now;
  if(inter[m] >max){
      max=inter[m];
    }
     if(inter[i]==-1){
        ans=0;
        
    }
//printf("%d\n",inter[i]);
  int fer[max+1];
  for (i=0;i<max+1;i++){
    switch (i){
      case 0:
        fer[i]=1;
        break;
      case 1:
        fer[i]=1;
        break;
      case 2:
       fer[i]=2;
       break;
      default:
        fer[i]=fer[i-2]+fer[i-1];
      break; 
     
    }
  //  printf("%d;%d\n",i,fer[i]);
  }
  for(i=0;i<m+1;i++){
    if(inter[i]==-1){
        ans=0;
    }else{
    ans*=fer[inter[i]];
        
    }
    ans%=1000000007;
   //  printf("%ld\n",ans);
  }
  printf("%ld\n",ans);
  return 0;
}