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)

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()
typedef long long ll;
// #include<C:\Users\J\Desktop\AC\athead.h>
#define showa(v) for(int iii = 0; iii < v.size(); iii++){ cout << v[iii] << " ";} cout << endl;
#define show(iiii) cout << iiii << endl; 
#define shows(iiiii) cout << iiiii << " ";
#define divs 1000000007
#include<iostream>
#include<string>
#include<vector>
#include<stdio.h>
#include <cmath>
#include <cstdlib>
#include <cassert>
#include <algorithm>
#include <set>
#include <deque>
#include <numeric>
#include<tuple>
#include <utility>
#include <map>
using namespace std;

int saiki(int n, vector<int> &memo){
    // showa(memo)
    int res = 0;
    if (memo[n] == -1){
        res += ((saiki(n-1, memo))+(saiki(n-2, memo)));
        memo[n] = res;
        return res;
    }
    else{
        return memo[n];
    }
}

int main(){
    int div = divs;
    int n, m;
    cin >> n >> m;
    vector<int> a(m+1);
    int hoji = -5;

    rep(i,m){
        cin >> a[i];
        if(hoji +1 == a[i] && hoji-1 < n){
            cout << 0;
            return 0;
        }
        hoji = a[i];
    }

    a[m] = n+1;
    int prev = 0;

    long long int res = 1;
    vector<int> memo(n+10);
    rep(j,n+10) memo[j] = -1;
    memo[0] = 0;
    memo[1] = 1;
    // cout << &memo << endl;

    rep(i,m+1){
        int haba = a[i] - prev;
        res = res%divs*saiki(haba, memo)%divs;
        prev = a[i]+1;
        // showa(memo)
    }

    cout << res;
}