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)

#include <bits/stdc++.h>
using namespace std;
static const long long MOD = 1000000007;

long long N, M;
set<long long> A;

void read() {
  cin >> N >> M;
  for (long long i = 0; i < M; i++) {
    long long a;
    cin >> a;
    A.insert(a);
  }
}

int main() {
  read();

  vector<long long> DP;
  DP.push_back(1);
  DP.push_back(A.find(1) == A.end() ? 1 : 0);

  for (long long i = 2; i <= N; i++) {
    long long size = static_cast<long long>(DP.size());
    long long back = DP.at(size - 1);
    long long tick = DP.at(size - 2);
    if (A.find(i) == A.end())
      DP.push_back((back + tick) % MOD);
    else
      DP.push_back(0);
  }

  cout << DP.back() << endl;
}