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: Time Limit Exceeded)

#include <bits/stdc++.h>

using namespace std;

#define ll long long
#define ld long double
#define P pair<ll, ll>

#define FORm(i, m) for (auto i = m.begin(); i != m.end(); i++)

#define all(v) v.begin(), v.end()

#define sortAl(v) sort(all(v))
#define sortAlr(v)          \
  sort(v.begin(), v.end()); \
  reverse(v.begin(), v.end())

#define cout(n) cout << fixed << setprecision(n)

#define rep(i, n) for (ll i = 0; i < n; i++)
#define REP(i, n) for (ll i = 1; i < n; i++)

template <class T>
inline void chmax(T &a, T b) {
  a = std::max(a, b);
}
template <class T>
inline void chmin(T &a, T b) {
  a = std::min(a, b);
}

#define mod (ll)(1e9 + 7)
#define INF LLONG_MAX

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  ll n, m;
  cin >> n >> m;

  set<ll> a;

  rep(i, m) {
    ll tmp;
    cin >> tmp;
    a.insert(tmp);
  }

  vector<ll> dp(n + 10);

  dp[0] = 1;
  if (m == 0 || *(a.begin()) != 1) {
    dp[1] = 1;
  }

  for (ll i = 2; i <= n; i++) {
    //段が壊れていなかったら
    if (find(all(a), i) == a.end()) {
      dp[i] = dp[i - 2] + dp[i - 1];
      dp[i] %= mod;
    }
  }

  cout << dp[n] << endl;

  return 0;
}