Problem:
Score : 500 points

Problem Statement

For an integer n not less than 0, let us define f(n) as follows:

f(n) = 1 (if n < 2)

f(n) = n f(n-2) (if n \geq 2)

Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).

Constraints

0 \leq N \leq 10^{18}

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of trailing zeros in the decimal notation of f(N).

Sample Input 1

12

Sample Output 1

1

f(12) = 12 × 10 × 8 × 6 × 4 × 2 = 46080, which has one trailing zero.

Sample Input 2

5

Sample Output 2

0

f(5) = 5 × 3 × 1 = 15, which has no trailing zeros.

Sample Input 3

1000000000000000000

Sample Output 3

124999999999999995

Solution in C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)


#include<stdio.h>

void fanction(double n, long int *cnt){
    int x = n;
    if(n < 2){
        return ;
    }else{
        while(x % 5 == 0){
            *cnt += 1;
            x /= 5;
        }
        return fanction(n - 2, cnt);
    }
}

int main(void){
    long int n, cnt = 0;
    scanf("%ld", &n);
    if(n % 2 == 1){
        printf("%d\n", 0);
        return 0;
    }

    fanction(n, &cnt);
    printf("%ld\n", cnt);
    return 0;
}