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>
//#include "mylib-cpp.h"

using namespace std;

int main(void){
    int n, m;
    cin >> n >> m;
    vector<int> dp(n);
    vector<int> a(n);
    for(int i=0;i<m;i++){
        int p;
        cin >> p;
        a[p-1] = 1;
    }
    for(int i=0;i<n;i++){
        if(i == 0){
            if(a[0] != 1) dp[0] = 1;
        }else if(i == 1){
            int f=1;
            if(a[1]) f=0;
            dp[1] = f*(dp[0] + 1);
        }else{
            int f=1;
            if(a[i]) f=0;
            dp[i] = f*(dp[i-2] + dp[i-1]) % 1000000007;
        }
    }
    cout << dp[n-1] << endl;
    return 0;
}