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>
#define Z 1000000007

int main(){
	int N,M,i,k;
	int end=0;
	long int result=1;
	scanf("%d %d",&N,&M);
	long int *F = (long int*)malloc(sizeof(long int)*N);
	int *a= (int*)malloc(sizeof(int)*M);
	int *s;
	if(M>1){
		s= (int*)malloc(sizeof(int)*M);
	}
	for(i=0;i<M;i++){
		scanf("%d",&a[i]);
		if(i>0 && M>1){
			s[i]=a[i]-a[i-1]-2;
			if(end==0){
				if(s[i]==-1)end=1;
			}
		}
	}
	if(end==1)printf("%d\n",0);
	else{
		F[0]=1,F[1]=1;
		for(i=2;i<N;i++){
			if(F[i-1]+F[i-2]>=Z){
				F[i]=(F[i-1]+F[i-2])%Z;
			}
			else F[i]=F[i-1]+F[i-2];
		}
		if(M>1){
		for(i=1;i<M;i++){
			k=s[i];
			result=result * F[k];
			if(result>=Z)result=result%Z;
		}
		}
		if(a[0]!=1){
			result=result * F[(a[0]-1)];
			if(result>=Z)result=result%Z;
		}
		result=result * F[(N-a[M-1]-1)];
		result=result%Z;
		printf("%ld\n",result);
	}
	return 0;
}