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 <bits/stdc++.h>
using namespace std;
const long long mod = 10e9 + 7;

int main(){
	long long N, M;
	cin >> N >> M;
	vector<long long> dp(N + 1);
	vector<bool> pos(N + 1, true);
	for(int i = 0; i < M; i++){
		long long n;
		cin >> n;
		pos[n] = false;
	}
	
	dp[0] = 1;
	if(pos[1]){
		dp[1] = 1;
	} else {
		dp[1] = 0;
	}
	
	for(int i = 2; i <= N; i++){
		if(pos[i]){
			dp[i] = dp[i - 1] + dp[i - 2];
			dp[i] %= mod;
		} else {
			dp[i] = 0;
		}
	}
	
	cout << dp[N] << endl;
	return 0;
}