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 <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>

#include <string>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <stack>
#include <deque>

#include <algorithm>

using namespace std;

int main(int argc, char *argv[]) {
	const int mod = 1000000007;
	
	vector<long long int> step(100001, 0);
	step[1] = 1;
	step[0] = 1;
	for (int i = 2; i <= 100000; i++) {
		step[i] = (step[i-1] + step[i-2]) % mod;
	}

	int start = 0;
	int n, m;
	vector<int> v;
	cin >> n;
	cin >> m;
	for (int i = 0; i < m; i++) {
		int t;
		cin >> t;
		v.push_back(t);
	}
	v.push_back(n+1);
	long long int res = 1;
	for (int i = 0; i < v.size(); i++) {
		if (start == v[i]) {
			cout << 0 << endl;
			return 0;
		}
		// cout << v[i] - 1 - start << endl;
		res *= step[v[i] - 1 - start];
		res %= mod;
		start = v[i] + 1;
	}
	cout << res << endl;
}