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>
#include <math.h>
#include <regex>
using namespace std;
typedef long long ll;
typedef long double ld;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
const ll mod = 1000000007;
const int INF = 1001001001;
const ll LINF = 1001001001001001001;
ld n , d;
vector<vector<ld>> x(n,vector<ld>(d));

int main(){
    ll n ; cin >> n;
    ll ans = 0;
    for(int i=0;i<=n;i++){
        if(i%3!=0 && i%5!=0){
            ans+=i;
        }
    }
    cout << ans << endl;
    return 0;
}