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 <string.h>

int main(){
	long long a;
    scanf("%lld",&a);
    long long sum=a*(a+1)/2;
    long long sumthree=(3+(a/3)*3)*(a/3)/2;
    long long sumfive=(5+(a/5)*5)*(a/5)/2;
    long long sumfiveteen=(15+(a/15)*15)*(a/15)/2;
    if(a>=15){
    	printf("%d\n",sum-sumthree-sumfive+sumfiveteen);
	}else if(a>=5){
    	printf("%d\n",sum-sumthree-sumfive);
	}else if(a>=3){
    	printf("%d\n",sum-sumthree);
	}else{
		if(a==2){
			printf("2\n");
		}else{
			printf("1");
		}
	}
    return 0;

}