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

//vector…可変長、データの大きさに合わせてサイズを変える
//list…双方向リスト、要素を挿入、削除の多いとき
//forward_list
//array…固定長配列
//set…集合
//multiset
//bitset…2進数、固定長ビット列
//map…連想配列、キーと値を関連付ける
//multimap
//deque
//stack…FILO
//queue…FIFO
//priority_queue…格納された最大値から順に取り出す
//pair…二つの値を一組
//tuple…複数の値を一組

template<typename T>
void remove(std::vector<T>& vector, char index) {
    vector.erase(vector.begin() + index);
}

template<class T>
T ma(std::vector<T> &v) {
    T maxi = 0;
    for(auto &i:v) maxi = max(maxi, i);
    return maxi;
}

template<class T>
T mi(std::vector<T> &v) {
    T mini = 0;
    for(auto &i:v) mini = min(mini, i);
    return mini;
}

template <class T>
T pow(T x, T y, T n = numeric_limits<T>::max()) {
    if(y == 0) return 1;
    if(y % 2 == 0) return pow((x * x) % n, y / 2);
    return (x * pow(x, y - 1)) % n;
}

#define rep(i, n, m) for (int i = n; i <= m; i++)
#define Rep(i, n, m) for (int i = m; i >= 1; i--)

typedef long long ll;
typedef unsigned long long ull;

/*
int gcd(int x, int y) {
    if (x < y) swap(x, y);
    return !y ? x : gcd(x % y, y);
}
*/
int main() {
    ull n, m;
    ull ans = 1;
    cin >> n >> m;
    vector<int> A(m+2);
    vector<int> result;
    A[0] = 0;
    for (int i = 1; i <= m; i++) cin >> A[i];
    A[m+1] = n;

    if (m >= 2) {
        rep(i, 1, m-2) {
            if (A[i+1] - (A[i]+1) == 0) {
                cout << 0 << endl;
                return 0;
            }
        }
    }
    cout << endl;

    rep(i, 0, m) {
        int x = A[i+1] - A[i] - 1;
        if (x) {
            ans = (ans * pow(2, x)) % 1000000007;
        }
    }

    cout << ans << endl;

    // int j = 0, cnt = 0;
    // rep(i, 1, n) {
    //     if (i == A[j]) {
    //         cnt = 0;
    //         flg[i] = cnt;
    //         j++;
    //     } else {
    //         flg[i] = ++cnt;
    //     }
    // }

    // flg[n] = 0;

    // rep(i, 1, n) {
    //     if (flg[i] == 0) {
    //         ans *= flg[i-1];
    //     }
    // }
    // cout << ans % 1000000007 << endl;
}