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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for( int i = 0 ; i < n ; i++ )
 
int main(){
  int mod = 1000000007;
  int N , M ;
  cin >> N >> M ;
  vector<int> a(M);
  vector<long long> dp(N+1);
  dp.at(0) = 1;
  dp.at(1) = 1;
  rep(i,M) cin >> a.at(i);
  rep(j,M) if( 1 == a.at(j) ) dp.at(1) = 0 ;
  for( int i = 1 ; i < N ; i++ ){
   dp.at(i+1) = ( dp.at(i) + dp.at(i-1) ) % mod  ;
   rep(j,M) if( i+1 == a.at(j) ) dp.at(i+1) = 0 ;
   //cout << "dp[" << i+1 << "]すなわち"<< i+1 << "段目に行く方法は" ;
   //cout << dp.at(i+1) << "通り" << endl;
  }
  cout << dp.at(N) << endl;
}