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;
#define all(v) ((v).begin()),((v).end())
#define ll long long
#define fastIO cout << fixed << setprecision(12), ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
double const EPS = 1e-12, PI = acos(-1);
const int N = 1e6 + 9, M = 5e3 + 9, OO = 1e9 + 7, MOD = 1e9 + 7;
const ll inf = 1e18;

bool la2[N];
int n, m, a;
ll mem[N];

ll dp(int step) {
  if(step == n)
    return 1;
  if(step > n)
    return 0;
  ll& ret = mem[step];
  if(~ret)
    return ret;
  ret = 0;
  if(!la2[step + 1])
    ret = ((ret % MOD) + (dp(step + 1) % MOD)) % MOD;
  if(!la2[step + 2])
    ret = ((ret % MOD) + (dp(step + 2) % MOD)) % MOD;
  return ret;
}

int main() {
  fastIO;
#ifdef LOCAL
  freopen("input.in", "rt", stdin);
#endif
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    cin >> a;
    la2[a] = 1;
  }
  memset(mem, -1, sizeof mem);
  cout << dp(0);
  return 0;
}