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>
using namespace std;
typedef long long ll;

#define REP(i, n) for (int i = 0; i < (int)(n); i++)
#define REPl(i, n) for (ll i = 0; i < (ll)(n); i++)
#define REPR(i, n) for(int i = n; i >= 0; i--)
#define REPRl(i, n) for(ll i = n; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i < (int)n; i++)
#define FORl(i, m, n) for(ll i = m; i < (ll)n; i++)
#define ALL(v) v.begin(), v.end()
ll gcd(ll a,ll b) {
  return b? gcd(b,a%b): a;  
}
ll lcm(ll a,ll b) {
  return a*b/gcd(a,b);
}
ll fibonacci(ll n){
  if(n==0){
    return 0;
  }else if(n==1){
    return 1;
  }else{
    return fibonacci(n-1)+fibonacci(n-2);
  }
}

int main() {
  ll n,m;cin>>n>>m;
  vector<ll> a(m+1);
  REPl(i,m){
    ll b;cin>>b;
    a[i]=b;
  }
  a[m]=n+1;
  vector<ll> p(m+1);
  p[0]=fibonacci(a[0]);
  FORl(i,1,m+1){
    p[i]=fibonacci(a[i]-a[i-1]-1);
  }
  ll ans=1;
  REPl(i,m+1){
    ans*=p[i];
  }
  cout << (ans%1000000007) << endl;
}