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>
#include <algorithm>
#include <cmath>

typedef long long ll;


void  calc(int num, std::vector<ll>& dp)
{
    for(int i=2; i<=num+1; ++i)
        dp[i] = dp[i-1] + dp[i-2];
}

int main() {

    int N, M;
    std::vector<ll> a;
    ll keyValue = 1000000007;

    std::cin >> N >> M;

    std::vector<ll> dp(N+2, 0);
    dp[0] = 0;
    dp[1] = 1;
    calc(N, dp);

    a.resize(M);
    for(int i=0; i<M; ++i)
        std::cin >> a[i];

    ll ans = 1;

    for(int i=0; i<a.size(); ++i)
    {
        if(i==0)
        {
            ll num = a[0];
            ans *= dp[num];
        }
        else
        {
            ll num = a[i] - a[i-1] -1;
            ans *= dp[num];
        }

        if(ans == 0)
        {
            std::cout << 0 << std::endl;
            return 0;
        }
    }

    if(M>0)
    {
        ll num = N - a[a.size()-1];
        ans *= dp[num];
    }
    else
    {
        std::cout << dp[N+1] % keyValue << std::endl;
        return 0;
    }

    std::cout << ans % keyValue << std::endl;


    return 0;
}