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 (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<stdio.h>
unsigned long long sum(int x){
  unsigned long long sum=0;
  return x/2*(x+1);
}
int main(){
  int N;
  scanf("%d",&N);
  int a=0,b=0,c=0;
  a=N/3;
  b=N/5;
  c=N/15;
  unsigned long long ans=0;
  ans=sum(N)-3*sum(a)-5*sum(b)+15*sum(c);
  printf("%d\n",ans);
  return 0;
}