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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
long long fib[100];
long long fibonatti(int n){
    long long a = 0;
    long long b = 1;
    for(int i=0; i<n-1; i++){
        long long c = b;
        b = a+b;
        a = c;
    }
    return b;
}
int main(){
    int n;
    scanf("%d", &n);
    int a[n+1];
    for(int i=0; i<n+1;i++){
        a[i]=1;
    }
    int m;
    scanf("%d", &m);
    for(int i=0; i<m;i++){
        int an;
        scanf("%d",&an);
        a[an]=0;
    }
    long long prod=1;
    int ck = 0;
    long long sum = 0;
    for(int i=0; i<n+1;i++){
        if(a[i]==0){
            if(ck==1){
                prod =0;
                break;
            }

                prod*=fibonatti(sum);
                sum=0;
            ck=1;
        }else{
            sum++;
            ck=0;
        }
    }
    prod*=fibonatti(sum);
    printf("%d\n", prod%1000000007);
    return 0;
}