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<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<time.h>
#define FOR(i,n) for(int i=0;i<n;i++)
#define PRN(n) printf("%d\n",n)
#define PRF(n) printf("%lf\n",n)
#define PRL(n) printf("%lld\n",n)
#define PRS(s) printf("%s\n",s)
#define PRC(c) printf("%c",c)
#define INI(n) scanf("%d",&n)
#define INL(n) scanf("%lld",&n)
#define INF 10000000000
#define mod 1000000007
#define X 100000
#define N 20
#define M 46
typedef long long int ll;
ll finv[X];
ll fac[X];
ll inv[X];
ll usort(const void *a, const void *b){return *(ll*)a-*(ll*)b;}
ll dsort(const void *a, const void *b){return *(ll*)b-*(ll*)a;}
ll ssort(const void *a, const void *b){return strcmp((char*)a,(char*)b);}
void Iswap(int *a,int *b){
  int t=*a;
  *a=*b;
  *b=t;
}
void Fswap(double *a,double *b){
  double t=*a;
  *a=*b;
  *b=t;
}
void Lswap(ll *a,ll *b){
  ll t=*a;
  *a=*b;
  *b=t;
}
ll min(ll a,ll b){
  if(a>b)return b;
  return a;
}
ll max(ll a,ll b){
  if(a>b)return a;
  return b;
}
ll z(ll a){return max(a,-a);}
ll gcd(ll a,ll b){if(!b){return a;}else{return gcd(b,a%b);}}
ll lcm(ll a,ll b){return a*b/gcd(a,b);}
ll kt(ll a){
  ll sum=0;
  while(a){
    a/=10;
    sum++;
  }
  return sum;
}
ll ks(ll a){
  ll sum=0;
  while(a){
    sum+=a%10;
    a/=10;
  }
  return sum;
}
void nCrReady(){
  fac[0]=fac[1]=1;
  finv[0]=finv[1]=1;
  inv[1]=1;
  for (ll i=2;i<X;i++){
    fac[i]=fac[i-1]*i%mod;
    inv[i]=mod-inv[mod%i]*(mod/i)%mod;
    finv[i]=finv[i-1]*inv[i]%mod;
  }
}
ll nCr(ll n,ll k){return fac[n]*(finv[k]*finv[n-k]%mod)%mod;}

int main(void){
  int n;
  scanf("%d",&n);
  ll ans=0;
  FOR(i,n+1)if(i%3&&i%5)ans+=i;
  PRL(ans);
  return 0;
}