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>
#include <limits.h>

long long fib1, fib2, fib3;
long long mod = 1000000007;

long long fib(int n, int length, int a[])
{
    int i;
    int flag = 0;
            
    fib3 = 1;
    if(a[1] == 1){
        fib2 = 0;
    }
    else{
        fib2 = 1;
    }
    
    for(i = 2; i <= n; ++i){
        flag = 0;        
        for(int j = 1; j <= length; j++){
            if(i == a[j]){
                flag = 1;
                break ;                
            }
        }
        if(flag == 1){
            fib1 = 0;
        }
        else{
            fib1 = (fib2 + fib3) ;
        
        }
        fib3 = fib2;
        fib2 = fib1;        
        
    }
    return fib1;
}

int main()
{
    int N, M;
    long long ans;
    
    scanf("%d%d", &N, &M);
    int a[M+1];
       
    
    for( int i = 1; i <= M; i++){
        scanf("%d", &a[i]);
    }
  
    ans = fib(N, M, a);    
    printf("%lld\n", ans%mod);
    
    return 0;
}