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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include "time.h"
#include "unistd.h"
#include "string.h"
#include <math.h>
int ans=0;
long long sort(long long d[],int NumberArrays,int up){
  int tmp;
  for (int i = 0; i < NumberArrays-1; i++) {
    if (up && d[i]<d[i+1]) {
      d[i]=tmp;
      d[i]=d[i+1];
      d[i+1]=tmp;
    }else if (d[i]>d[i+1]) {
      d[i]=tmp;
      d[i]=d[i+1];
      d[i+1]=tmp;
    }
  }
}
int main(int argc, char* argv[]){
	long long int a = 0, b = 0, c = 0,d[10000];
  char f[3];
	scanf("%lld",&a);
  for (long int i = 1; i < a+1; i++) {
    if (i%3!=0 && i%5!=0) {
      b+=i;
    }
  }
  printf("%lld\n",b);
}