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>
long long step_variation(long long n){
  long long onebf = 0;
  long long twobf = 0;
  long long result = 0;
  long long oldresult = 0;
  long long i = 0;
  if(n==0){
    return 1;
  }else if(n==1){
    return 1;
  }else if(n==2){
    return 2;
  }else{
    onebf=2;
    twobf=1;
    for(i=0;i<n-2;i++){
      result = (onebf + twobf)%1000000007;
      twobf = onebf;
      onebf = result;
    }
    return result%1000000007;
  }
}

int main(){
  int n=0;
  int m =0;
  int a[100000] ={};
  int b[100000] = {};
  long long i = 0;
  long long total = 0;
  int successable = 0;
  scanf("%d %d",&n,&m);
  
  for(i=0;i<m;i++){
    scanf("%d",&a[i]);
  }
  
  //printf("%ld/",step_variation(5));
  for(i=0;i<m;i++){
    if(i==0){
      b[i] = a[i]-1;
    }else{
      b[i] = a[i] - a[i-1] -2;
    }
  }
  b[m] = (n-2)-a[m-1]+1;
  
  /*for(i=0;i<=m;i++){
    printf("%ld/",b[i]);
  }*/
  total = 1;
  for(i=0;i<m+1;i++){
    total = (total * step_variation(b[i]))%1000000007;
  }
  for(i=0;i<m-1;i++){
    if(a[i]+1 == a[i+1]){
        total = 0;
    }
  }
  printf("%lld",total);
  
  return 0;
}