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: Accepted)

#include <stdio.h>

int main(void) {
    long long N, x, p, i, j;
    scanf("%lld", &N);
    if (N%2==1) {
        printf("0");
        return 0;
    }
    x=0; j=1;
    while (1) {
        p=2;
        for (i=1; i<=j; i++) p*=5;
        if (N/p!=0) x+=N/p;
        else break;
        j++;
    }
    printf("%lld", x);
    return 0;
}