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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<stdio.h>

long long call(long long n) {
    if (n < 2) {
        return 1;
    }
    else {
        return n * call(n - 2);
    }
}


int main(void) {
    long long a = 0, b = 0, i;
    scanf("%lld", &i);
    a = call(i);
    while (a > 1) {
        if (a % 10 == 0) {
            b++;
        }
        else if (a % 10 != 0) {
            break;
        }
        a = a / 10;
    }
    printf("%d", b);
    return 0;
}