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)

#include<iostream>
#include<math.h>

using namespace std;

// 指定する項のフィボナッチ
long long getFibonacci(int target){
    long long f0 = 0, f1 = 1;
    long long fn = 1;

    for(long long i = 1; i < target; i++){
        fn = f0 + f1;
        f0 = f1;
        f1 = fn;
    }

    return fn;
}

int main(){

    int N, M;

    cin >> N >> M;

    // 番兵
    int a[M + 1];

    for(int i = 0; i < M; i++){
        cin >> a[i];
    }
    a[M] = N + 1;

    long long ans = 1;
    int prev = -1;

    int diff;

    for(int i = 0; i < M + 1; i++){
        diff = a[i] - prev;
        if(diff == 1){
            cout << 0 << endl;
            return 0;
        }
        if(diff > 1){
            ans = ans * getFibonacci(diff - 1) % 1000000007;
        }
        prev = a[i];
    }

    cout << ans << endl;

    return 0;
}