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: Runtime Error)

#include<iostream>
#include<string>
#include<vector>
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<numeric>
#include<iomanip>
#include<deque>
#include<tuple>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<cstdint>
#include<unordered_map>
#include<sstream>
#include<boost/multiprecision/cpp_int.hpp>

#define rep(i, n) for(int i = 0; i < ( int)(n); i++)
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define vi vector<int>
#define vll vector<ll>
#define all(x) (x).begin(),(x).end()
#define F first
#define S second
#define pb push_back

namespace mp = boost::multiprecision;
using cpp_int = mp::cpp_int;
using ll = long long;
using namespace std;

ll M = 1000000007;

typedef pair<int, int> P;

bool ps1(const P& element1, const P& element2)
{//first昇順
	return element1.first < element2.first;
}
bool ps2(const P& element1, const P& element2)
{//first降順
	return element1.first > element2.first;
}
bool ps3(const P& element1, const P& element2)
{//second昇順
	return element1.second < element2.second;
}
bool ps4(const P& element1, const P& element2)
{//second降順
	return element1.second > element2.second;
}

ll combination(ll n, ll r) {
	if (r * 2 > n) r = n - r;
	ll dividend = 1;
	ll divisor = 1;
	for (unsigned int i = 1; i <= r; ++i) {
		dividend *= (n - i + 1);
		divisor *= i;
	}
	return dividend / divisor;
}

int main()
{
	//cout << combination(18, 2) << endl;
	int n, m;
	cin >> n >> m;
	vector<bool>step(n+1, true);
	rep(i, m) {
		int a;
		cin >> a;
		step[a] = false;
	}
	cpp_int count = 1;
	queue<ll>k;
	int tmp = -1;
	rep(i, n+1) {
		if (step[i] == true) {
			tmp++;
		}
		else if (step[i] == false && step[i + 1] == false) {
			count = 0;
			break;
		}
		else if (step[i] == false && step[i + 1] == true) {
			if (tmp > 0) {
			//	cout << tmp << endl;

				int t = 0;
				ll x = 0;
				while (2*t<= tmp) {
					int s = tmp - 2*t + t;
					x += combination(s, t);
					t++;
				}
				k.push(x);
				tmp = -1;
			}
			tmp = -1;

			
		}
		count %= M;
	}
	//cout << tmp << endl;
	if (tmp > 0) {
	//	cout << tmp << endl;
		int t = 0;
		ll x = 0;
		while (2*t<= tmp) {
			int s = tmp - 2*t + t;
			x += combination(s, t);
			t++;
		}
		k.push(x);
	}
	tmp = -1;
	count %= M;
	while (k.size() != 0) {
		count *= k.front();
		count %= M;
		k.pop();
	}
	cout << count << endl;
}