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: Time Limit Exceeded)

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int vector_finder(std::vector<int> vec, int number) {
	auto itr = std::find(vec.begin(), vec.end(), number);
	size_t index = std::distance(vec.begin(), itr);
	if (index != vec.size()) { // 発見できたとき
		return 1;
	}
	else { // 発見できなかったとき
		return 0;
	}
}
int main() {
	int n, m;
	cin >> n >> m;
	vector<int> a(m),v(n);
	for (int i = 0; i < m; i++) {
		cin >> a[i];
	}
	v[0] = 1;
	v[1] = 2;
	if (vector_finder(a, 1)) {
		v[0] = 0;
		v[1] = 1;
	}
	if (vector_finder(a, 2))v[1] = 0;
	for (int i = 2; i < n; i++) {
		if (vector_finder(a, i+1)) {
			v[i] = 0;
		}
		else v[i] = (v[i - 1] + v[i - 2]) % 1000000007;
	}
	cout << v[n-1]<<endl;
	return 0;
}