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>

#define rep(i, n) for(int i = 0, i##_len = (n); i < i##_len; ++i)
#define repp(i, m, n) for(int i = m, i##_len = (n); i < i##_len; ++i)
#define all(x) (x).begin(), (x).end()
#define clr(ar, val) memset(ar, val, sizeof(ar))
template <class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return 1; } return 0; }
template <class T> bool chmin(T &a, const T &b) { if (b < a) { a = b; return 1; } return 0; }
int gcd(int a,int b){return b?gcd(b,a%b):a;}
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair <int,int> P;
typedef long double ld;

int main(void)
{
	int n, m, a[100001] = {}, y[100001] = {0, 1};
	cin >> n >> m;
	rep(i, m) cin >> a[i];
	int next = a[0], it = 0;
	rep(i, n){
		bool danger = false;
		if(i + 1 == next){
			danger = true;
			if(it + 1 < m) next = a[++it];
		}
		if(!danger)
			if(i == 0)
				y[2] = 1;
			else
				y[i + 2] = (y[i] + y[i + 1]) % 1000000007;
		else
			y[i + 2] = 0;
	}
	// rep(i, n) cout << y[i + 2] << " ";
	cout << y[n + 1] << endl;
	return 0;
}