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>
#define in_range(x, y, r, c) (x < r && y < c && x >= 0 && y >= 0)
using namespace std;
const int mod = 1e9+7;
bool no[100100];
int memo[100100];

int ups(int v){
	if (v == 0) return 1;
	if (v < 1) return 0;
	if (no[v]) return 0;
	if (memo[v] != -1) return memo[v];
	return memo[v] = ups(v-1) + ups(v-2);
}

int main(){
	int n, m, aux;
	cin >> n >> m;
	memset(no, false, sizeof no);
	memset(memo, -1, sizeof memo);
	for (int i=0; i<m; i++){
		cin >> aux;
		no[aux] = true;
	}
	cout << ups(n) << endl;
    return 0;
}