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: Wrong Answer)

#define rep(i,n) REP(i,0,n)
#define REP(i, a, b) for(int i = a; i < (int)(b); i++)
#include <bits/stdc++.h>

typedef long long ll;
using namespace std;

ll DIV = (ll)1e9+7;
ll Fibo[(ll)1e5+100];
ll GetFibo (ll id) {
   if (!Fibo[id]) {
      Fibo[id] = (GetFibo(id-2) + GetFibo(id-1))%DIV;
   }
   return Fibo[id];
}

int main(void) {
   ll N,M;
   cin>>N>>M;
   ll A[M];
   rep(i,M) {
      cin>>A[i];
   }
   rep(i,N) Fibo[i] = 0;
   Fibo[0] = 1;
   Fibo[1] = 1;
   Fibo[2] = 2;

   ll ans = 1;
   if (M == 0) ans = GetFibo(N-0);
   else {
      ans *= GetFibo(A[0]-1-0)%DIV;
      ans *= GetFibo(N-A[M-1]-1)%DIV;
      rep(i,M-1) {
         if (A[i+1]-A[i]<=1) ans *= 0;
         else ans *= GetFibo(A[i+1] - A[i]-2)%DIV;
      }
   }
   cout<<ans%DIV<<endl;

   return 0;
}