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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define INF 1e9
#define ALL(v) v.begin(), v.end()
#define REP(i, n) for(int i = 0; i < n; i++)
#define max(v) *max_element(v.begin(), v.end())
#define min(v) *min_element(v.begin(), v.end())
using namespace std;
 
int main(void){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, m;
    cin>>n>>m;
    int s;
    int a[m];
    REP(i,m){
        cin>>s;
        a[i] = s;
    }
    unsigned long long dp[100100] = {0};
    int key;
    dp[0] = 1;
    for(int i = 1; i < n+1; i++){
        if(i == 1 and a[0] != 1){
            dp[1] += 1;
        }
        else{
            key = 0;
            for(int j = 0; j < m; j++){
                if(i == a[j]){
                    key = 1;
                    break;
                }
            }
            if(key == 0){
                dp[i] += dp[i-1];
                dp[i] += dp[i-2];
                dp[i] %= 1000000007;
            }
        }
    }
    /*
    REP(i, n+1){
        cout<<i<<dp[i]<<endl;
    }
    */
    cout<<dp[n]%1000000007<<endl;
    return 0;
}