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>
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define RFOR(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define REP(i,n) FOR(i,0,(n))
#define REPS(i,n) FOR(i,1,(n)+1)
#define RREP(i,n) RFOR(i,0,(n))
#define ALL(x) (x).begin(),(x).end()
#define RALL(x) (x).rbegin(),(x).rend()
#define DEBUG(x)  cout << #x << " = " << (x) << endl;
#define SORT(x) sort(ALL(x));
#define RSORT(x) sort(RALL(x));
#define SUM(x) accumulate(ALL(x),0);
#define FI first
#define SE second
#define PB push_back
#define EB emplace_back
#define SZ(x) (int)(x).size()
#define BN(x) ((1<<x)-1)
#define DUP(x,y) (((x)+(y)-1)/(y))
using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using pii = pair<int, int>;
using tiii = tuple<int, int, int>;
const ll mod = 1e9+7;
const int INF = (1<<30)-1;
const ll INFLL = (1LL<<62)-1;
const int dx[4] = {1,0,-1,0};
const int dy[4] = {0,1,0,-1};

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; }
 
ll gcd(ll a, ll b) { return b ? gcd(b,a%b) : a;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}


int main() {
  int N, M;
  cin >> N >> M;
  vector<int> A(M);
  bool flg = false;
  for (int i = 0; i < M; ++i) {
    cin >> A[i];
    if (0 < i) {
      if (A[i - 1] == A[i] - 1) {
        flg = true;
        break;
      }
    }
  }
  if (flg) {
    cout << 0 << endl;
  } else {
    vector<ll> dp(N + 1);
    for (int i = 0; i <= N; ++i) {
      if (i <= 1) {
        dp[i] = 1;
        continue;
      }
      auto skip = find(A.begin(), A.end(), i);
      auto skip2 = find(A.begin(), A.end(), i - 1);
      auto skip3 = find(A.begin(), A.end(), i - 2);
      if (skip == A.end()) {
        if (skip2 != A.end() || skip3 != A.end()) {
          dp[i] = dp[i - 1];
        } else {
          dp[i] = dp[i - 1] + dp[i - 2];
        }
      } else {
        dp[i] = dp[i - 1];
      }
    }
    ll result = dp[N] % 1000000007;
    cout << result << endl;
  }
}