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: Runtime Error)

#include<bits/stdc++.h>
typedef long long ll;
#define rep(i,x) for(int i=0;i<x;i++)

using namespace std;
ll N = 1000000007;


int main(){
    int dp[100000] = {0};
	dp[0] = 0;
	dp[1] = 1;
	for(int i = 2; i < 100000; i++){
		dp[i] = dp[i-1] + dp[i-2];
	}
    int n,m;
    cin >> n >> m;
    vector<ll> a(m);
    vector<ll> b(m+1);
    vector<ll> c(m+1);
    ll ans = 1;
    rep(i,m){
        ll temp;
        //scanf("%llu",&temp);
        cin >> temp;
        a.at(i) = temp;
        if(i > 0 && a.at(i-1) + 1 == a.at(i)){
            cout << 0 << endl;
            return 0;
        }
        if(i == 0) b.at(i) = a.at(i);
        else b.at(i) = a.at(i) - a.at(i-1) - 1; 
    }
    b.at(m) = n - a.at(m-1);
    rep(i,m+1){
        c.at(i) = dp[b.at(i)];
        // if(ans > N) ans  = ans * c.at(i) % N;
         ans *= c.at(i);
    //    cout << c.at(i) << endl;
    }
    if(ans > N) ans %= N;
    cout << ans << endl;
    // cout <<dp[5];
    return 0;
    
}