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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define ld long double
#define mod 1000000007LL
#define eps 1e-13
#define PI 3.141592653589793238L
#define INF 1000000011
#define INFLL 10000000011LL
#define vi vector<int>
#define vll vector<long long>
#define pii pair<int, int>
#define pll pair<long long, long long>
#define pil pair<int, long long>
#define pli pair<long long, int>
#define mp make_pair
#define F first
#define S second
#define pb push_back
#define fo(i,a,n) for(i = (a); i < (n); i++)
#define gtl(x) getline(cin, (x))
#define flsh fflush(stdout)
#define sws ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define gcd __gcd
#define all(a) (a).begin(), (a).end()
#define sz(a) (int)((a).size())
#define io_file freopen("D:/Coding Problems/Contest/input_file.in", "r", stdin); freopen("D:/Coding Problems/Contest/output_file.out", "w", stdout)

ll modx(ll Base, ll exponent)
{
	ll ans = 1;
	if(Base == 1)
		return Base;
	while(exponent)
	{
		if(exponent & 1)
			ans = (ans * Base)%mod;
		Base = (Base * Base)%mod;
		exponent = exponent >> 1;
	}
	return ans;
}

ll inmodx(ll num)
{
	return (modx(num, mod-2LL));
}

bool cmp(pair < vi, int > a, pair < vi, int > b)//true for a before b
{
	if(a.F.empty())
		return 1;
	if(b.F.empty())
		return 0;
	int i;
	fo(i,0,min(sz(a.F),sz(b.F)))
	{
		if(a.F[i] < b.F[i])
			return 1;
		else if(a.F[i] > b.F[i])
			return 0;
	}
	return (sz(a.F) <= sz(b.F));
}

const int N = (1e5) + 9;
const int M = (N<<2) + 9;
const int LOGN = ((int)log2(N)) + 3;
const int LOGM = ((int)log2(M)) + 3;
const int BUCK = 900;
const int SQRT = BUCK+9;

bool mark[N];
ll dp[N];

void solve()
{
	int n, m, i, a;
	cin >> n >> m;
	fo(i,0,n+1)
		mark[i] = 1;
	fo(i,1,m+1)
	{
		cin >> a;
		mark[a] = 0;
	}
	dp[0] = 1;
	dp[1] = (mark[1]) ? 1 : 0;
	fo(i,1,n+1)
	{
		dp[i] = (dp[i-1] + dp[i-2])%mod;
		if(!mark[i])
			dp[i] = 0;
	}
	cout << dp[n] << '\n';
	return;
}

int main()
{
	sws;
	clock_t clk;
	clk = clock();
	// io_file;
	// srand (time(NULL));

	//Code here
	int t = 1, cs;
	cout << fixed << setprecision(15);
	// cin >> t;
	fo(cs,1,t+1)
	{
		// cout << "Case #" << cs << ": ";
		solve();
	}
	// Code ends here

	clk = clock() - clk;
	cerr << fixed << setprecision(6) << "Time: " << ((double)clk)/CLOCKS_PER_SEC << "\n";
	return 0;
}