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: Runtime Error)

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


long long Fibonacci(int x){
	long long dp[1000] = {0};
	dp[0] = 0;
	dp[1] = 1;
	for(int i = 2; i < x; i++){
		dp[i] = dp[i-1] + dp[i-2];
	}
	return dp[x];
}

int main() {
  int N,M;cin >> N >> M;
  long long t,s;
  int a[100010];bool can = true;
  for(int i = 0; i < M; i++){
  cin >> a[i];
  }
  long long ans = 1;
  for(int i = 1; i < M; i++) {
  if(a[i]-a[i-1] == 1){
    cout << 0 << endl;
    can = false; break;
  }
  ans *= Fibonacci(a[i]-a[i-1])%1000000007;
  }
  
  if(can){
    cout << (ans*Fibonacci(a[0])*Fibonacci(N-a[M-1]))%1000000007 << endl;
  }
  
}