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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
#define REP(i,b,e) for(ll i=(ll)b;i<(ll)e;i++)
#define rep0(i,n) REP(i,0ll,n)
#define rep1(i,n) REP(i,1ll,n+1)
const ll longinf=1ll<<60;
const int inf=(1<<30)-1;

const ll mod=1e9+7;
int main(){
      ll n,m;
      cin>>n>>m;
      ll a[m];
      rep0(i,m){
            cin>>a[i];
      }
      bool ng[n+1];
      fill(ng,ng+n+1,false);
      rep0(i,m){
            ng[a[i]]=true;
      }
      ll ptn[n+1];
      fill(ptn,ptn+n+1,0);
      ptn[0]=1;
      if(!ng[1])ptn[1]=1;
      REP(i,2,n+1){
            if(!ng[i]){
                  ptn[i]=(ptn[i-1]+ptn[i-2])%mod;
            }
      }
      cout<<ptn[n]<<endl;

      return 0;
}