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: Accepted)


#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse2")

    /*
    ||||||||||||||||||||||||||||||||||||||||||||||||
       |      |||         |      ||||    ||||  |   |
      | |     |   |      | |     |   |  |      |   |
     |   |    |    |    |   |    |   |  |      |   |
    |||||||   |    |   |||||||   |||     |||   |||||
    |     |   |   |    |     |   | |        |  |   |
    |     |   |||      |     |   |  ||  ||||   |   |
    ||||||||||||||||||||||||||||||||||||||||||||||||

    */
    using namespace std;
    //#include "testlib.h"
    #define ff first
    #define ss second
    #define mp make_pair
    #define all(v) v.begin(),v.end()
    #define int long long
    #define ll long long
    #define M 1000000007

    #define mii  map<ll ,ll >
    #define msi  map<string,ll >
    #define mis  map<ll int, string>
    #define rep(a,b)    for(ll i=a;i<b;i++)
    #define rep0(n)    for(ll i=0;i<n;i++)
    #define repi(i,a,b) for(ll i=a;i<b;i++)
    #define pb push_back
    #define vi vector<ll>
    #define mp make_pair
    #define vs vector<string>
    #define ppb pop_back
    #define endl '\n'
    #define asdf ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    #define r0 return 0;
    #define FORD(i, a, b) for (int i = (int) (a); i >= (int) (b); --i)
    #define FORE(it, c) for (__typeof((c).begin()) it = (c).begin(); it != (c).end(); ++it)
    #define inputoutput freopen("input.txt", "r", stdin);freopen("output.txt", "w", stdout);
    #define input freopen("input.txt", "r", stdin);
    #define Set(a, s) 4(a, s, sizeof (a))
    #define FOR repi
    #define pii pair<int,int>
    //#define float long double
    ll max(ll a, ll b) { return (a > b)? a : b;}
    int min(int a, int b) { return (a < b)? a : b;}


     int solve()
     {
        int n;
        cin>>n;
        int m;
        cin>>m; vi ans(n+1,1);
        vi a(n+1,0); rep0(m) {int t;cin>>t;a[t]=1;ans[t]=0;}
       
        rep(1,n+1){
            if(a[i]) continue;
            ans[i]=ans[i-1];
            if(i>=2) ans[i]=(ans[i]+ans[i-2])%M;        
     
        }
        cout<<ans[n];
        }
    signed main()
    {   
        asdf
        int t=1;
        //cin>>t;
        while(t--)
        {
            solve();
        }     
        
}