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: Accepted)

#include <stdio.h>
int main(){
	int n,m,i,a,dp[100002]={0};
	scanf("%d%d",&n,&m);
	for(i=0;i<m;i++){scanf("%d",&a);dp[a]=-1;}
	dp[0]=1;
	dp[1]=(dp[1]==-1?-1:1);
	for(i=2;i<=n;i++){
		if(dp[i-1]==-1&&dp[i-2]==-1){printf("0");return 0;}
		else if(dp[i]<0)continue;
		else if(dp[i-1]==-1)dp[i]=dp[i-2];
		else if(dp[i-2]==-1)dp[i]=dp[i-1];
		else dp[i]=dp[i-1]+dp[i-2];
		dp[i]%=1000000007;
	}
	printf("%d",dp[n]);
}