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: Wrong Answer)

#include<bits/stdc++.h>
#include<boost/multiprecision/cpp_int.hpp>
using namespace std;
using namespace boost::multiprecision;
#define BEGIN ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define END return EXIT_SUCCESS
#define rep(I,N) for(auto I=0;I<(N);++I)
#define up(I,A,B) for(auto I=(A);I<=(B);++I)
#define dw(I,A,B) for(auto I=(A);I>=(B);--I)
#define all(C) (C).begin(),(C).end()
#define rall(C) (C).rbegin(),(C).rend()
#define ft first
#define sd second
#define mp make_pair
#define mt make_tuple
#define pf push_front
#define pb push_back
#define pt pop_front
#define pk pop_back
#define eb emplace_back
#define lb lower_bound
#define ub upper_bound
#define rs resize
// #define int cpp_int
template<class T>inline bool chmax(T& a,T b){if(a<b){a=b;return true;}return false;}
template<class T>inline bool chmin(T& a,T b){if(a>b){a=b;return true;}return false;}
template<class T>inline void                                      in(T &p){cin>>p;}
template<class T1,class T2>inline void                            in(T1 &p1,T2 &p2){cin>>p1>>p2;}
template<class T1,class T2,class T3>inline void                   in(T1 &p1,T2 &p2,T3 &p3){cin>>p1>>p2>>p3;}
template<class T1,class T2,class T3,class T4>inline void          in(T1 &p1,T2 &p2,T3 &p3,T4 &p4){cin>>p1>>p2>>p3>>p4;}
template<class T1,class T2,class T3,class T4,class T5>inline void in(T1 &p1,T2 &p2,T3 &p3,T4 &p4,T5 &p5){cin>>p1>>p2>>p3>>p4>>p5;}
template<class T>inline void                             out(T p){cout<<p<<endl;}
template<class T1,class T2>inline void                   out(T1 p1,T2 p2){cout<<p1<<" "<<p2<<endl;}
template<class T1,class T2,class T3>inline void          out(T1 p1,T2 p2,T3 p3){cout<<p1<<" "<<p2<<" "<<p3<<endl;}
template<class T1,class T2,class T3,class T4>inline void out(T1 p1,T2 p2,T3 p3,T4 p4){cout<<p1<<" "<<p2<<" "<<p3<<" "<<p4<<endl;}
inline                                              void out(double p){printf("%.10f\n",p);}
cpp_int dp[1000005],mod=(long)1e9+7;
inline void solve(void){
  long N,M;
  in(N,M);
  long a;
  rep(i,M){
    in(a);
    dp[a]=-1;
  }
  dp[1]=dp[1]!=-1?1:-1;
  dp[2]=dp[1]==-1?1:2;
  up(i,3,N){
    if(dp[i]==-1)continue;
    if(dp[i-1]!=-1)dp[i]+=dp[i-1],dp[i]%=mod;
    if(dp[i-2]!=-1)dp[i]+=dp[i-2],dp[i]%=mod;
  }
  out(dp[N]);
}
int main(int argc,char**argv){
  BEGIN;
  solve();
  END;
}