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 f[1000000];


long long fib(int a,int b,long long k){
  if(a == 0){
    return 0;
  }
  else if(a == b-1){
    return k;
  }
  
  f[0] = 1;
  
  k += f[b-1];
  f[b] = k;
  b ++;
  fib(a,b,k);
}


int main(void){
  int n , m;
  scanf("%d %d",&n,&m);
  
  int a[m + 2];
  a[0] = 0 , a[m+1] = n;
  for(int i = 1;i <= m;i ++){
    scanf("%d",&a[i]);
  }
  
  long long ans = 1; 
  
  for(int i = 0;i <= m;i ++){
    int b = a[i+1] - a[i] -1;
    if(a[i+1] == 1){
      ans += 1;
      continue;
    }
      ans *= fib(b,1,0);
      if(ans >= 1000000007){
        ans %= 1000000007;
    }
  }
  
  printf("%lld",ans);
  
  return 0;
}