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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn = 1e5+5;
const ll mod = 1000000007;
int dp[maxn];
int vis[maxn];
int main()
{
    int n,m;
    memset (dp,0,sizeof(dp));
    memset (vis,0,sizeof(vis));
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++){
        int temp;
        scanf("%d",&temp);
        vis[temp]=1;
    }
    int falg=0;
    for(int i=2;i<=n;i++){
        if(vis[i]&&vis[i-1]){
            falg=1;
            break;
        }
    }
    if(falg){
        cout<<0<<endl;
    }
    else{
        dp[0]=1;dp[1]=1;
        for(int i=2;i<=n;i++){
            if(vis[i-1]){
                dp[i]=dp[i-2];
            }
            else if(vis[i-2]){
                dp[i]=dp[i-1];
            }
            else{
                dp[i]=(dp[i-1]+dp[i-2])%mod;
            }
        }
        printf("%d\n",dp[n]);
    }
}