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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <limits.h>
#include <time.h>
int main(){
    long long i,n,ans=0,five[25]={};
    scanf("%lld",&n);
    five[0]=10;
    if(n%2==1){
        printf("0");
        return 0;
    }
    for(i=1; i<25; i++){
        five[i]=5*five[i-1];
    }
    for(i=0; i<25; i++){
        ans+=n/five[i];
    }
    printf("%lld",ans);
    return 0;
}