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>
#include<stdlib.h>

int main(void){
  int n,m;
  int i,j;

  scanf("%d%d",&n,&m);
  int a[m];
  unsigned long long int b[n];
  for(i=0;i<n;i++)
  b[i]=3;


  for(i=0;i<m;i++){
    scanf("%d",&a[i]);
    b[a[i]-1]=0;
    if(i>0&&(a[i]+1==a[i-1])){
      printf("%d",0);
      return 0;
    }
  }


  //ここから先は二つ連続で穴が来ることはないから。

  if(b[0]==0)
  b[1]=1;
  else if(b[1]==0)
  b[0]=1;
  else{
    b[0]=1;
    b[1]=2;
  }

  for(i=2;i<n;i++){
    if(b[i]!=0)
    b[i]=b[i-1]+b[i-2];
    while(b[i]>1000000007)
    b[i]=b[i]-1000000007;
  }
  b[n-1]=b[n-1]%1000000007;
  //for(i=0;i<n;i++)
  //printf("%ddanme:%lld\n",i+1,b[i]);
  printf("%lld",b[n-1]);
  return 0;
}