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 main ()
{
    int n,i;
    long long x=0;
    scanf("%d", &n);
    int arr[n];
    for(i=1; i<=n; i++){
        if(i%3==0 && i%5==0){arr[i]=n+3;}
        else if(i%3==0){arr[i]=n+2;}
        else if(i%5==0){arr[i]=n+1;}
        else {arr[i]=i;
              x=x+i;}
    }
    printf("%lld\n", x);
    for(i=1 ;i<=n; i++){
        if(arr[i]==n+3){printf("FizzBuzz,");}
        else if(arr[i]==n+2){printf("Fizz,");}
        else if(arr[i]==n+1){printf("Buzz,");}
        else {printf("%d,", i);}
    }
}