Count occurrences 2

Count occurrences 2
Given a non-negative int n, compute recursively (no loops) the count of the occurrences of 8 as a digit, except that an 8 with another 8 immediately to its left counts double, so 8818 yields 4.

Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6), while divide (/) by 10 removes the rightmost digit (126 / 10 is 12).

Input
On the first line you will be given n.

Output
On the only output line you should print the count of 8s.

Constraints
n>=0

Sample tests
Input
8
Output
1
Input
818
Output
2

import java.util.Scanner;

public class CountOccurrences2 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int input = Integer.parseInt(scanner.nextLine());
        System.out.println(numberOfOccurrences(input));
    }

    public static int numberOfOccurrences(int n){
        int count = 0;
        if(n < 10 && n != 8){
            return count;
        }
        if(n%10==8){
            count++;
            if(n/10 % 10 == 8){
                count++;
            }
        }
        return count + numberOfOccurrences(n/10);
    }
}