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;

typedef long long ll;
const ll mod = 1e9 + 7;
const int MAXN = 1e5 + 1;
int n, m;
ll dp[MAXN];
bool visit[MAXN];

ll solve(int pos) {
	
	if (pos == n) return 1;
	if (visit[pos] || pos>n) return 0;

	ll& ret = dp[pos];

	if (ret != -1) return ret;

	ret = 0;

	ret += solve(pos + 1)%mod;
	ret += solve(pos + 2)%mod;

	return ret;
}

int main() {

	cin >> n >> m;

	for (int i = 0; i < m; i++) {
		int val;
		cin >> val;
		visit[val] = true;
	}

	memset(dp, -1, sizeof(dp));

	cout << solve(0) << "\n";

	return 0;
}