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;

int fiv( int x){
    long long current = 1;
    long long next = 1;
    for ( int i = 0; i < x; i++){
        long tmp = current;
        current = next;
        next += tmp;
    }
    return current;
}

int main(){
    int n, m;
    cin >> n >> m;
    vector<int> a(m);
    for (int i = 0 ; i < m; i++){
        cin >> a.at(i);
    }
    long long count = fiv(a.at(0) - 1 );
    count *= fiv( n - a.at( m - 1 ) - 1);

    int flag = 1;
    for (int i = 1 ; i < m; i++){
        if (a.at(i) - a.at(i - 1) == 1){
            count = 0;
            flag = 0;
            break;
        }
    }

    if (flag == 0){
        cout << count << endl;
        return 0;
    }

    for ( int i = 0 ; i < m - 1 ; i++){
        if ( a.at( i + 1 ) - a.at( i ) != 2 ) {
            count *= fiv(a.at(i + 1) - a.at(i) - 2);
        }
    }

    cout << count % 1000000007 << endl;
}