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 <vector>

using namespace std;

long long calc_comb(int p, int r);

long long calc_routes(int stair_count, int broken_count, vector<int> broken_stairs);

// p >= r
long long calc_comb(int p, int r)
{
    long long mul = 1;
    if(p - r > r){
        r = p - r;
    }

    for(int i = p; i > r; i--){
        mul *= i;
    }

    for(int i = 2; i <= p - r; i++){
        mul /= i;
    }

    return mul;
}

long long calc_routes(int stair_count, int broken_count, vector<int> broken_stairs)
{
    long long route_sum = 1;

    int split_pointer = 0;
    vector<int>::iterator itr;
    for(itr = broken_stairs.begin(); itr != broken_stairs.end(); itr++){
        if(itr + 1 != broken_stairs.end()){
            if(*(itr + 1) - *itr == 1){
                return 0;
            }
        }
        
        int i = *itr - split_pointer - 1;
        int j = 0;
        long long part_sum = 0;
        while(i >= j){
            part_sum += calc_comb(i, j);

            i--;
            j++;
        }

        route_sum *= part_sum;
        route_sum %= 1000000007;
        // ? 
        split_pointer = *itr + 1;
    }

    int i = stair_count - split_pointer;
    int j = 0;
    long long part_sum = 0;
    while(i >= j){
        part_sum += calc_comb(i, j);

        i--;
        j++;
    }

    route_sum *= part_sum;

    return route_sum;
}

int main(void)
{
    int stair_count;
    int broken_count;

    vector<int> broken_stairs;

    cin >> stair_count >> broken_count;

    for(int i = 0; i < broken_count; i++){
        int bs;
        cin >> bs;
        broken_stairs.push_back(bs);
    }


    cout << calc_routes(stair_count, broken_count, broken_stairs) % 1000000007 << endl;

    return 0;
}