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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using llong = long long;

void fnInput(int& rnStair, vector<int>& rvnBroken)
{
  int nBrokenSiz;
  cin >> rnStair >> nBrokenSiz;
  rvnBroken.resize(nBrokenSiz);

  for (int& rnBroken : rvnBroken)
    cin >> rnBroken;
}

bool fnUseFulStairSet(int nStair, vector<int>& rvnUseful, const vector<int>& cnrvnBroken)
{
  for (int i = 0; i < cnrvnBroken.size(); i++)
    if (i == 0)
      if (cnrvnBroken[i] == 1);
      else
        rvnUseful.push_back( cnrvnBroken[i] - 1 );
    else
      if (cnrvnBroken[i] - cnrvnBroken[i - 1] == 1)
        return false;
      else
        rvnUseful.push_back( (cnrvnBroken[i] - 1) - (cnrvnBroken[i - 1] + 1) );

  if (cnrvnBroken.size())
    rvnUseful.push_back( nStair - (cnrvnBroken[cnrvnBroken.size() - 1] + 1) );
  else
    rvnUseful.push_back( nStair );
  return true;
}

void fnDPWaySet(int nStair, vector<int>& rvnDPWays)
{
  rvnDPWays.resize(nStair + 1);
  rvnDPWays[0] = 1;

  for (int i = 1; i < rvnDPWays.size(); i++)
  {
    rvnDPWays[i] += rvnDPWays[i - 1];
    if (i > 1)
      rvnDPWays[i] += rvnDPWays[i - 2];
  }
}

int fnNumOfWays(int nStair, const vector<int>& cnrvnBroken)
{
  vector<int> vnDPWays;
  fnDPWaySet(nStair, vnDPWays);

  vector<int> vnUseful;
  if (!fnUseFulStairSet(nStair, vnUseful, cnrvnBroken))
    return 0;

  llong nNumOfWays = 1;

  for (int nUseful : vnUseful)
    nNumOfWays *= vnDPWays[ nUseful ];
  
  return nNumOfWays % 1000000007;
}
  
int main()
{
  int nStair;
  vector<int> vnBroken;

  fnInput(nStair, vnBroken);
  cout << fnNumOfWays(nStair, vnBroken) << endl;

  return 0;
}