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)

#ifndef _GLIBCXX_NO_ASSERT
#include <cassert>
#endif
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <ciso646>
#include <climits>
#include <clocale>
#include <cmath>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>

#if __cplusplus >= 201103L
#include <ccomplex>
#include <cfenv>
#include <cinttypes>
#include <cstdbool>
#include <cstdint>
#include <ctgmath>
#include <cwchar>
#include <cwctype>
#endif

// C++
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>

#if __cplusplus >= 201103L
#include <array>
#include <atomic>
#include <chrono>
#include <condition_variable>
#include <forward_list>
#include <future>
#include <initializer_list>
#include <mutex>
#include <random>
#include <ratio>
#include <regex>
#include <scoped_allocator>
#include <system_error>
#include <thread>
#include <tuple>
#include <typeindex>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#endif




#define int long long
#define rng(i,l,r) for(size_t i=(l);i<(r);++i)
#define rep(i,n) rng(i,0,n)
#define gnr(i,l,r) for(size_t i=(r)-1;i>=(l);i--)
#define per(i,b) gnr(i,0,b)
#define ALL(obj) (obj).begin(), (obj).end()
#define rALL(obj) (obj).rbegin(), (obj).rend()

using namespace std;
//using ll = long long;


const int MOD = 1000000007;
const int INF = 1e18;

const int dx4[4] = {1, 0, -1, 0};
const int dy4[4] = {0, 1, 0, -1};
const int dx8[8] = {1,1,0,-1,-1,-1,0,1};
const int dy8[8] = {0,1,1,1,0,-1,-1,-1};

const double pi = acos(-1);



template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }


using Graph = vector<vector<int>>;



void solve(){
    
    // remove the bottom 3 lines when you submit this code.
    std::ifstream in("sample.txt");
    std::cin.rdbuf(in.rdbuf());

    cin.tie(0);
    ios::sync_with_stdio(false);
    
    int n,m; cin >> n >> m;
    vector<int> a(m+100);
    
    for (int i = 0; i < m; i++) cin >> a[i];
    
    //calc fib
    int fib[100010];
    for (int i = 0; i <= 100010; i++) {
        if(i==0) {
            fib[i] = 1;
        }else if(i==1) {
            fib[i] = 1;
        }else{
            fib[i] = fib[i-1] + fib[i-2];
            fib[i] %= MOD;
        }
    }
    
    
    
    int ans = 1;
    for (int im  = 0; im <= m; im++) {
        int da = 0;
        if(im==0)      da = a[im] - (-1)    - 2;
        else if(im==m) da = (n+1) - a[im-1] - 2;
        else           da = a[im] - a[im-1] - 2;
        
        
        (da>=0)?ans *= fib[da]%MOD:ans *= 0;
        
        ans %= MOD;
    }
    
    cout << ans << endl;
    
    
    
    
    
    
    
}



signed main() {
    solve();
    return 0;
}