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 <cstdio>
#include <algorithm>
#define repi(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,a) repi(i,0,a)
#define all(a) (a).begin(), (a).end()

using ll = long long;
constexpr ll mod = 1000000007;

ll N, M;
ll a[100010];
ll dp[100010];

int main()
{
  scanf( "%lld%lld", &N, &M );
  rep( i, M )
    scanf( "%lld", a+i );

  int p = 0, q = 0;

  if( a[0] == 1 )
    ++q;

  dp[0] = 1;

  rep( i, N )
  {
    if( i+1 <= N && a[p] != i+1 )
      dp[i+1] = (dp[i+1]+dp[i]) % mod;
    if( a[p] == i+1 )
      ++p;
    
    if( i+2 <= N && a[q] != i+2 )
      dp[i+2] = (dp[i+2]+dp[i]) % mod;
    if( a[q] == i+2 )
      ++q;
  }

  printf( "%lld\n", dp[N] );

  return 0;
}