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>
#define ll long long
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define deb(x) cout<<#x<<"="<<x<<endl;
#define endl '\n'
#define M 1000000007
#define int long long
#define max_n 1000000

using namespace std;

void solve()
{
	ll n;
	cin>>n;
	ll ans=0;
	for (int i = 1; i <= n; ++i)
	{
		if(i%3==0 || i%5==0)
		{}
		else ans+=i;
	}
	cout<<ans<<endl;
}

int32_t main() 
{
	IOS	
	ll T=1;
	// cin>>T;
	for(ll i=1;i<=T;++i)
	{		
		// cout<<"Case #"<<i<<": ";
    	solve();
	}
	return 0;
}