Problem:
Score : 200 points

Problem Statement

Let us define the FizzBuzz sequence a_1,a_2,... as follows:

If both 3 and 5 divides i, a_i=\mbox{FizzBuzz}.

If the above does not hold but 3 divides i, a_i=\mbox{Fizz}.

If none of the above holds but 5 divides i, a_i=\mbox{Buzz}.

If none of the above holds, a_i=i.

Find the sum of all numbers among the first N terms of the FizzBuzz sequence.

Constraints

1 \leq N \leq 10^6

Input

Input is given from Standard Input in the following format:

N

Output

Print the sum of all numbers among the first N terms of the FizzBuzz sequence.

Sample Input 1

15

Sample Output 1

60

The first 15 terms of the FizzBuzz sequence are:

1,2,\mbox{Fizz},4,\mbox{Buzz},\mbox{Fizz},7,8,\mbox{Fizz},\mbox{Buzz},11,\mbox{Fizz},13,14,\mbox{FizzBuzz}

Among them, numbers are 1,2,4,7,8,11,13,14, and the sum of them is 60.

Sample Input 2

1000000

Sample Output 2

266666333332

Watch out for overflow.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
long  long n[1000005]={0};
int main() 
{
    long long i,z,j=0;
   
    long long  sum=0;
    scanf("%lld",&z);
    for(i=1;i<=z;i++)
    {
    	if(i%3==0||i%5==0||i==z)
    	    continue;
	    n[j]=i;
	    j++;
	}
	for(i=0;i<=z;i++)
	{
		sum =sum+n[i];
	}
	  
	printf("%lld",sum);
    
	return 0;
}