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: Wrong Answer)

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<cmath>	

using namespace std;

long long calc(long long k) {
	long long r = 1;
	for (long long i = 1; i <= k; ++i) {
		r *= i;
	}
	return r;
}

int main() {
	vector<long long> A;
	vector<long long> step;
	long long N;
	long long M;
	long long preStep = 0;
	cin >> N;
	cin >> M;

	long long sumCount = 1;

	for (long long i = 0; i <= M; ++i) {
		int stepCount = 0;
		vector<long long> count;
		count.push_back(1);
		count.push_back(1);
		long long step;
		if (i == M) {
			step = N + 1;
		}
		else {
			cin >> step;
		}
		if (step == preStep) {
			cout << 0 << endl;
			return 0;
		}
		for (long long j = preStep; j < step; ++j) {
			if (stepCount <= 1) {
				++stepCount;
			}
			else {
				count.push_back((count[count.size() - 2] + count[count.size() - 1]) % 1000000007);
			}
		}
		sumCount *= count[count.size() - 1] % 1000000007;
		preStep = step + 1;

	}
	cout << sumCount % 1000000007 << endl;


	return 0;
}