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)

/*!enjoy karo yaar!*/

//CF,CC,AtC,SPOJ: hp1999
//HE: hemant269
//HR: hemant2132

#include<bits/stdc++.h>
using namespace std;

#define     int             long long int
#define     fast()          ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define     all(x)          (x).begin(),(x).end()
#define     rz(x)           resize(x)
#define     mem(a,b)        memset(a,b,sizeof(a))
#define     sz(x)           ((int)(x.size()))
#define     pb              push_back
#define     pf              push_front
#define     pob             pop_back
#define     pof             pop_front
#define     ins             insert
#define     vi              vector<int>
#define     pii             pair<int,int>
#define     mii             map<int,int>
#define     F               first
#define     S               second
#define     makep           make_pair
#define     maket           make_tuple
#define     remax(a,b)      a=max(a,b)
#define     remin(a,b)      a=min(a,b)
#define     bitcount(x)     __builtin_popcountll(x)
#define     iceil(n,x)      (((n)+(x)-1)/(x))
#define     gcd(a,b)        (__gcd((a),(b)))
#define     lcm(a,b)        (((a)*(b))/gcd((a),(b)))
#define     dbug(x)         cout<<#x<<": "<<(x)<<"\n"
#define     flush           fflush(stdout)
#define     show(x)         for(auto zz:x)cout<<zz<<" ";cout<<"\n";
#define     show2(x)        for(auto zz:x)cout<<zz.F<<" "<<zz.S<<"\n";

typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;

const ld pi=acos(-1);
const ll inf=1e18,M=1e9+7;
const int N=1;

void solve()
{
    int n;
    cin>>n;

    int sum=0;
    for(int i=1;i<=n;++i)
    {
        if(i%5!=0 && i%3!=0)
            sum+=i;
    }

    cout<<sum;
}

int32_t main()
{
    fast();

    int t=1;
    //cin>>t;
    for(int z=1;z<=t;++z)
    {
        solve();
        //cout<<"\n";
    }

    return 0;
}