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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <cstdio>
#include <map>
#include <algorithm>
#include <queue>
using namespace std;
int N,M;
const int mod=1000000007;
int dp[100001];
int main(){
	cin >> N >> M;
	dp[0]=1;
	dp[1]=1;
	for(int i=1;i<=M;i++){
		int a;
		cin >> a;
		dp[a]=-1;
	}
		
	for(int i=2;i<=N;i++){
		if(dp[i]!=-1){
			if(dp[i-1]!=-1)dp[i]+=dp[i-1];
			if(dp[i-2]!=-1)dp[i]+=dp[i-2];
		}
		dp[i]%=mod;
	}
	printf("%d\n",dp[N]%mod);
	return 0;
}