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>
using namespace std;
int main()
{
	int n, m, nn[100100], dp[100801];
	cin >> n >> m;
	for (int i = 0; i <= n; i++)
	{
		dp[i] = 0;
		nn[i] = 0;
	}
	for (int i = 0; i < m; i++)
	{
		cin >> nn[i];
	}
	dp[0] = 1;
	if (nn[0] != 1)
	{
		dp[1] = 1;
	}
	int c = 0;
	for (int i = 2; i <= n; i++)
	{
		if (nn[c] == i - 2)
		{
			if (nn[c + 1] == i - 1)
			{
				if (nn[c + 2] == i)
				{
					c += 3;
				}
				else
				{
					c += 2;
				}
			}
			else
			{
				if (nn[c + 1] == i)
				{
					c += 2;
				}
				else
				{
					dp[i] = dp[i - 1];
					c++;
				}
			}
		}
		else
		{
			if (nn[c] == i - 1)
			{
				if (nn[c + 1] == i)
				{
					c += 2;
				}
				else
				{
					c += 1;
					dp[i] = dp[i - 2];
				}
			}
			else
			{
				if (nn[c] == i)
				{
					c += 1;
				}
				else
				{
					dp[i] = (dp[i - 1] + dp[i - 2]) % 1000000007;
				}
			}
		}
	}
	cout << dp[n] << endl;
	return 0;
}