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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <bits/stdc++.h>
#include <string>

using namespace std;
typedef long long ll;
const long double EPS = 1e-9;
const ll mod = 1e9 + 7;

#define rep(i, n) for(int i = 0; i < n ; i++ )
#define For(i, a, b) for(int i = (a); i < (b) ; i++ )
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()

void init(ll patterns[], ll N){
    patterns[0] = 1;
    patterns[1] = 1;

    For(i, 2, N + 1){
        patterns[i] = (patterns[i - 1] + patterns[i - 2]) % mod;
    }
}

int main(void){
    ll N, M;
    cin >> N >> M;

    ll a[M];
    ll patterns[N + 1];
    ll answer = 1;

    init(patterns, N);

    rep(i, M){
        cin >> a[i];
    }

    For(i, -1, M){
        ll idx;

        if(i == -1){
            idx = a[0] - 1;
        }else if(i == M - 1){
            idx = N - (a[M - 1] + 1);
        }else{
            idx = (a[i + 1] - 1) - (a[i] + 1);
        }
        
        if (idx < 0){
            answer = 0;
            break;
        }

        answer = (answer * patterns[idx]) % mod;
    }

    cout << answer << endl;

    return 0;
}