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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;

long long fib[101]={0};

long long test(int n){	// フィボナッチ数列をメモする関数
	fib[0]=1;
	fib[1]=1;

	if(fib[n])return fib[n];
	else{
		fib[n]=test(n-1)+test(n-2);
		return fib[n];
	}
}



int main(){
	long long int now=0,step=0,ans=1;
	int x;
	x=1000000007;
	long long a;
	for(int i=0;i<101;i++)a=test(i);


	int n,m;	cin>>n>>m;
	if(m==0)cout<<fib[n]<<endl;
	else{
		int broken[m];
		for(int i=0;i<m;i++)cin>>broken[i];


		int j=0;
		while(now<n){
			if(now+1!=broken[j]){
				now++;
				++step;
			}else{
				if(now+2==broken[min(j+1,m-1)]){
					ans=0;
					break;
				}else{
					ans *= fib[step];
					ans%=x;
					//				cout<<"NOW: "<<now<<endl;
					//				cout<<"STEP: "<<step<<endl;
					//				cout<<"FIB["<<step<<"]= "<<fib[step]<<endl;
					//				cout<<endl;

					j++;

					now+=2;
					step=0;
				}
			}

		}
	}

	ans *= fib[step];
	ans%=x;

	if(m)cout<<ans<<endl;

	return 0;
}