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: Time Limit Exceeded)

#include<stdio.h>
#include<stdlib.h>

int n, m, a[100001],b[100001];

long way(int j){
	long count=0;
	if(j==0) return 0;
	else if(j==1){
		if(b[j]==0) return 1;
		else return 0;
	}else{
		if(b[j-1]==0) count += way(j-1);
		if(b[j-2]==0) count += way(j-2);
		return count;
	}

}

int main(){
	int i;
	fscanf(stdin,"%d %d", &n, &m);
	for(i=0;i<n+1;i++){
		b[i] = 0;
	}
	for(i=0;i<m;i++){
		fscanf(stdin,"%d", &a[i]);
		b[a[i]+1]=1;
	}
	
	printf("%d\n",way(n+1)%1000000007);
	return 0;
}