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 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int mul(int x,int sum)
{
	int a=0,bs=0,i;
	for(i=1;a<=sum;i++)
  {
	  bs+=a;
	  a=i*x;
  }
	return bs;
}
int main ()
{
  int i,sum=0,a=0,bs=0;
  char s[7];
  scanf("%s",s);
  for(i=0;s[i];i++)
  sum=sum*10+s[i];
  bs=mul(3,sum)+mul(5,sum)-mul(15,sum);
  sum=(sum+1)*sum/2-bs;
  printf("%d",sum);
  return 0;
}