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;
typedef long long ll;
typedef pair<int, int> pii;
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep2(i, a, b) for (int i = (a); i < (b); ++i)
#define MAX 128
#define fst first
#define sec second
#define pb push_back
#define mp make_pair
#define ALL(x) (x).begin(), (x).end()
#define debug(x) cerr << #x << ": " << x << '\n'
#define bit(n) (1LL << (n))
#define int long long
const double PI = 3.141593;
const int MOD = (int)1e9 + 7;
const int INF = 100100100;
const double EPS = 1e-9;
template <class T>
bool chmax(T &a, const T &b){if (a < b){a = b;return 1;}return 0;}
template <class T>bool chmin(T &a, const T &b){if (b < a){a = b;return 1;}return 0;}
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; } //最大公約数
ll lcm(ll a, ll b) { return a * b / gcd(a, b); }     //最小公倍数
ll combination(ll n, ll r){if (r * 2 > n)r = n - r;ll dividend = 1;ll divisor = 1;for (ll i = 1; i <= r; ++i){dividend *= (n - i + 1);divisor *= i;}return dividend / divisor;}


signed main()
{
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);

  int n;
  cin >> n;
  int fizzbuzz, fizz, buzz;
  fizzbuzz = n / 15;
  fizz = n / 3;
  buzz = n / 5;
  int sum = n * (1 + n) / 2;
  int ans = sum - 3 * (1 + fizz) * fizz / 2 - 5 * (1 + buzz) * buzz / 2 + 15 * (1 + fizzbuzz) * fizzbuzz / 2;
  cout << ans << endl;
}