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;
using ll =long long;
using db = double;

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

#define che(a,string) cout<<"//"<<string <<"=="<<(a)<<"//" <<endl;
#define SORT(a) sort((a).begin(),(a).end())
#define rSORT(a) reverse((a).begin(),(a).end())
#define For(i, a, b)    for(int i = (a) ; i < (b) ; ++i)
#define rep(i, n)       For(i, 0, n)
#define leftunique(v) SORT(v);(v).erase(unique((v).begin(),(v).end()),(v).end())
//leftuniqueは配列から同じ要素のものを取り除く 例 4 6 2 1 3 4 1 2 なら 1 2 3 4 6 になる
bool IsInt(double a){int b = a/1;if(a==b){return true;}else {return false;}}
bool coY() {cout <<"Yes"<<endl;}
bool coN(){cout <<"No"<<endl;}
const int mod = 1e9 + 7;
const ll INF = 1LL << 60;

ll n;
ll m;
ll dp[10010];


//ここから書き始める
int main()
{
	//dpの初期化
	rep(i,10010){
		dp[i] = 1;
	}
	cin >> n >>m;
	rep(i,m){
		ll broken_stair;
		cin >> broken_stair;
		dp[broken_stair] = 0;
	}
	For(i,2,n+1){
		if(dp[i]!=0){
			dp[i] = dp[i-1] + dp[i-2];
		}
	}
	cout <<dp[n]%mod <<endl;
}