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)

//author:luckYrat(twitter:@luckYrat_)
//<ここに一言>

//せんげん！
#include <iostream>

#include <cmath>
#include <algorithm>
#include <iomanip>

#include <string>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <map>
#include <bitset>

#include <cctype>
#include <utility>
#include <climits>


//なまえくーかん！
using namespace std;
using ll = long long;
using P = pair<ll,ll>;

//てーすう！
const int mod = 1000000007;
const int inf = (1<<30)-1;
const ll linf = (1LL<<62LL)-1;
const double EPS = (1e-10);

//でふぁいん！
#define anyfill(n,s) setw(n) << setfill(s)
#define loop(s) for(int i = 0; s > i; i++)
#define rep(i,q) for(int i = 0; (q) > i; i++)
#define repp(i,n,q) for(int i = n; (q) > i; i++)
#define dep(i,q) for(int i = (q); 0 < i; i--)

//みじかく！
#define pb push_back
#define fir first
#define scn second
#define ednl endl

//いぇすのー！
#define YesNo(a) (a?"Yes":"No")
#define YESNO(a) (a?"YES":"NO")
#define yesno(a) (a?"yes":"no")

//きんぼーnほーこー！！
P ar4[4] = {{0,1},{0,-1},{1,0},{-1,0}};
P ar8[8] = {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
/*
確認ポイント
cout << fixed << setprecision(n) << 小数計算//n桁の小数表記になる

計算量は変わらないが楽できるシリーズ
min(max)_element(iter,iter)で一番小さい(大きい)値のポインタが帰ってくる
count(iter,iter,int)でintがiterからiterの間にいくつあったかを取得できる
*/
__attribute__((constructor))
void initial() {
 cin.tie(0);
 ios::sync_with_stdio(false);
}
int main(){
  int n;cin>>n;
  long long ans =0 ;
  for(int i = 1; n >= i; i++){
    if(i%3 && i%5){
      ans += i;
    }
  }
  cout << ans << endl;
}