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>
#include<stdlib.h>
int main(){
  //  int a;
    long long int a,total,total3,total5,total15;
    int x;
    char s[10];
    // 整数の入力
  fgets(s,10,stdin);
  sscanf(s,"%lld\n",&a);
  total = (long long int)((1+a)*a/2);
  
  x = a/3;
  //printf("%d ",x);
  total3 = (long long int)((3+3*x)*x/2);
  x=a/5;
  //printf("%d ",x);
  total5 = (long long int)((5+5*x)*x/2);
  x=a/15;
  //printf("%d ",x);
  total15 = (long long int)((15+15*x)*x/2);
  total = total - total3 - total5 + total15;
  printf("%lld",total);
  
}