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>
#include <time.h>

int main(void){
	long i;
	long long ans=1;
	
	long n,m;
	scanf("%ld%ld",&n,&m);
	
	long a[m+2];
	a[0]=0;
	a[m+1]=n;
	for(i=1;i<=m;i++){
		scanf("%ld",&a[i]);
	}
	
	long d[m+1];
	for(i=0;i<=m;i++){
		d[i]=a[i+1]-a[i]-1;
	}
	if(a[1]==1){
		d[0]=1;
	}
	if(a[m]==n-1){
		d[m]==1;
	}
	
	long fib[n];
	fib[0]=0,fib[1]=1,fib[2]=2,fib[3]=3;
	for(i=4;i<=n;i++){
		fib[i]=(fib[i-1]+fib[1-2])%1000000007;
	}
	
	for(i=0;i<=m;i++){
		ans=(long long)(ans*fib[d[i]])%1000000007;
	}
	
	printf("%lld\n",ans);
	
	return 0;
}