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: Accepted)

///============================================================================
/// Name        : Faisal.cpp
/// Author      : Faisal Ahmed
/// Version     :
/// Copyright   : Your copyright notice
/// Description : C++, Ansi-style
///============================================================================

#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;
typedef long double ld;

bool f[100010];
ll arr[100010];

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL); cout.tie(NULL);

	ll n, m, x;

	cin >> n >> m;

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

	arr[0] = 1;
	for(int i = 0; i < n; i++){
        if(!f[i + 1]){
            arr[i + 1] += arr[i];
            arr[i + 1] = arr[i + 1] % 1000000007;
        }
        if(!f[i + 2]){
            arr[i + 2] += arr[i];
            arr[i + 2] = arr[i + 2] % 1000000007;
        }
	}

    cout << arr[n] << endl;

	return 0;
}