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>

int main(){

  long n;
  long i;
  long ans[1000000];
  long sum = 0;
  
  scanf("%ld",&n);
  
  for(i = 0; i <= n; i++){
  	if(i % 15 == 0 || i % 5 == 0 || i % 3 == 0){
      ans[i] = 0;
      //printf("%d\n", ans[i]);
    }else{
      ans[i] = i;
      //printf("%d\n", ans[i]);
    }
  }
    
    for(i = 0; i <= n; i++){
      sum += ans[i];
    }
   
    printf("%ld", sum);
  
  return 0;
}