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 <iostream>
#include <iomanip>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <deque>
#include <algorithm>
#include <cmath>
#include <cstring>

using namespace std;

#define REP(i,n) for(int i = 0; i < n; i++)
#define RREP(i,n) for(int i = n - 1; i >= 0; i--)

#define ALL(v) v.begin(),v.end()
#define pb push_back
#define mp make_pair
#define F first
#define S second

#define UNIQUE(v) do{sort(v.begin(), v.end()); v.erase( unique(v.begin(), v.end()), v.end() );} while(0)


#define y0 y3487465
#define y1 y8687969

#define m0(x) memset(x,0,sizeof(x))
#define m1(x) memset(x,63,sizeof(x))

typedef long long ll;
typedef pair<int,int> pi;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pi> vpi;
typedef vector<ll> vll;
typedef vector<vll> vvll;

double EPS = 1e-9;
int INFi = 1000000005;
long long INFll = 1000000000000000005ll;
double PI = acos(-1);
int dirx[8] = {-1,0,0,1,-1,-1,1,1};
int diry[8] = {0,1,-1,0,-1,1,-1,1};
ll MOD = 1000000007;

int N, M;
const int MAX_M = 100000;
int a[MAX_M];

int main(){
  cin >> N >> M;
  REP(i, M) cin >> a[i];

  REP(i, M - 1) {
    if(a[i + 1] == a[i] + 1) {
      cout << 0 << endl;
      return 0;
    }
  }

  ll dp[N];
  dp[0] = 1;
  dp[1] = 1;
  REP(i, N - 2) {
    dp[i + 2] = (dp[i + 1] + dp[i]) % MOD;
  }

  ll ans = dp[a[0] - 1];
  REP(i, M - 1) {
    ans = ans * dp[a[i + 1] - a[i] - 2] % MOD;
  }
  ans = ans * dp[N - 1 - a[M - 1]] % MOD;

  cout << ans << endl;

  return 0;
}