Problem:
Score : 200 points

Problem Statement

Takahashi loves gold coins. He gains 1000 happiness points for each 500-yen coin he has and gains 5 happiness points for each 5-yen coin he has. (Yen is the currency of Japan.)

Takahashi has X yen. If he exchanges his money so that he will gain the most happiness points, how many happiness points will he earn?

(We assume that there are six kinds of coins available: 500-yen, 100-yen, 50-yen, 10-yen, 5-yen, and 1-yen coins.)

Constraints

0 \leq X \leq 10^9

X is an integer.

Input

Input is given from Standard Input in the following format:

X

Output

Print the maximum number of happiness points that can be earned.

Sample Input 1

1024

Sample Output 1

2020

By exchanging his money so that he gets two 500-yen coins and four 5-yen coins, he gains 2020 happiness points, which is the maximum number of happiness points that can be earned.

Sample Input 2

0

Sample Output 2

0

He is penniless - or yenless.

Sample Input 3

1000000000

Sample Output 3

2000000000

He is a billionaire - in yen.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
 
// Type alias
using ll=long long;
using ld=double;
 
using pi=pair<int,int>;
using pll=pair<ll,ll>;
using pld=pair<ld,ld>;
using ti3=tuple<int, int, int>;
 
using vi=vector<int>;
using vll=vector<ll>;
using vld=vector<ld>;
using vpi=vector<pi>;
using vpll=vector<ll>;
using vpld=vector<ld>;
using vti3=vector<ti3>;
 
// Constants
const double EPS = 1e-9;
const int inf = numeric_limits<int>::max();
 
// Macro
#define rep(i,n) for(int i=0;i<int(n);++i)
#define rrep(i,n) for(int i=int(n-1);i>=0;--i)
#define REP(i,a,b) for(int i=int(a);i<int(b);++i)
#define RREP(i,a,b) for(int i=int(b-1);i>=int(a);--i)
#define SHOW(a) cout << #a << " = " << a << endl
#define ARR(a,n) for(int i=0;i<int(n);++i) cout << #a << "[" << i << "]" << " = " << a[i] << endl
 
#define ALL(a) a.begin(),a.end()
 
// Funtcions
ll pow(ll base, ll i, ll mod){
  ll a = 1;
  while(i){
    if (i & 1) {
      a *= base;
      a %= mod;
    }
    base *= base;
    base %= mod;
    i >>= 1;
  }
  return a;
}
ll gcd(ll a, ll b){
  while(b){
    ll c = a % b;
    a = b;
    b = c;
  }
  return a;
}
ll lcm(ll a, ll b){
  return a / gcd(a, b) * b;
}
 
 
void solve(){
  int n;
  cin >> n;
  int res = (n / 500) * 1000 + n % 500 / 5 * 5;
  cout << res << endl;
}
 
int main(){
  // FastIO
  //ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}