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<bits/stdc++.h>
using namespace std;
#define ll long long
#define lvector vector<ll>
#define lque queue<ll>
#define lpque priority_queue<ll>
#define dlpque priority_queue<ll,lvector,greater<ll>>
#define P pair<ll,ll>
#define ALL(a) (a).begin(),(a).end()
#define rep(i,n) for(ll (i)=0; (i)<(n); ++(i))
#define print(a) cout << (a) << endl

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ll n,ans=0;cin>>n;
  for(ll i=1;i<=n;++i) if(i%3!=0&&i%5!=0) ans+=i;
  print(ans);
  return 0;
}