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;
 
#define int long long
#define dub long double
#define FOR(i, a, b) for (int i=(a); i<(b); ++i)
#define ROF(i, a, b) for (int i=(a); i>(b); --i)
#define CASET int _T; cin >> _T; FOR(caset, 0, _T)
#define pii pair<int,int>
#define t3i tuple<int,int,int>
#define mii map<int,int>
#define vi vector<int>
#define vb vector<bool>
#define vs vector<string>
#define vvi vector<vi>
#define vvb vector<vb>
#define vvvi vector<vvi>
#define vvvb vector<vvb>
#define vpii vector<pii>
#define vt3i vector<t3i>
#define vvpii vector<vpii>
#define sq(x) (x)*(x)
#define inv(x) powa(x, MOD-2)
#define ALL(x) (x).begin(), (x).end()
#define RALL(x) (x).rbegin(), (x).rend()
#define SIZE(x) (signed)(x).size()
#define SUM(x) accumulate(ALL(x), 0LL)
 
long long INF = 1LL<<60;
long long MOD = 998244353;
 
namespace output {
    void pr(short x) { cout << x; }
    void pr(signed x) { cout << x; }
    void pr(long x) { cout << x; }
    void pr(long long x) { cout << x; }
    void pr(unsigned short x) { cout << x; }
    void pr(unsigned x) { cout << x; }
    void pr(unsigned long x) { cout << x; }
    void pr(unsigned long long x) { cout << x; }
    void pr(float x) { cout << x; }
    void pr(double x) { cout << x; }
    void pr(long double x) { cout << x; }
    void pr(char x) { cout << x; }
    void pr(const char* x) { cout << x; }
    void pr(const string& x) { cout << x; }
    void pr(bool x) { cout << x; }
    template<size_t sz> void pr(const bitset<sz>& x) { cout << x; }
    template<class T> void pr(const complex<T>& x) { cout << x; }
    
    template<class T1, class T2> void pr(const pair<T1,T2>& x);
    template<class T1, class T2, class T3> void pr(const tuple<T1,T2,T3>& x);
    template<class T> void pr(const T& x);
    
    template<class T, class... Ts> void pr(const T& t, const Ts&... ts) { pr(t); pr(ts...); }
    template<class T1, class T2> void pr(const pair<T1,T2>& x) { pr("(",x.first,", ",x.second,")"); }
    template<class T1, class T2, class T3> void pr(const tuple<T1,T2,T3>& x) { pr("(",get<0>(x),", ",get<1>(x),", ",get<2>(x),")"); }
    template<class T> void pr(const T& x) { 
        pr("["); // const iterator needed for vector<bool>
        bool fst = 1; for (const auto& a: x) pr(!fst?", ":"",a), fst = 0; 
        pr("]");
    }
    
    void print() { pr("\n"); } // print w/ spaces
    template<class T, class... Ts> void print(const T& t, const Ts&... ts) { pr(t); if (sizeof...(ts)) pr(" "); print(ts...); }
}
using namespace output;
 
template <class T1, class T2> istream &operator>>(istream &is, pair<T1,T2> &x) { is >> x.first >> x.second; return is; }
template <class T1, class T2, class T3> istream &operator>>(istream &is, tuple<T1,T2,T3> &x) { is >> get<0>(x) >> get<1>(x) >> get<2>(x); return is; }
template <class T> istream &operator>>(istream &is, vector<T> &v) { for (auto &x : v) is >> x; return is; }
int powa(int base, int exp) {
    int res = 1;
    while (exp) {
        if (exp&1) res = res*base % MOD;
        base = base*base % MOD;
        exp >>= 1;
    }
    return res;
}
bool mini(int &a, int b) { return b < a ? a = b, 1 : 0; }
bool maxi(int &a, int b) { return b > a ? a = b, 1 : 0; }

main() {
    ios::sync_with_stdio(0), cin.tie(0);
    
    int n;
    cin >> n;

    int a = n / 500;
    int res = a * 1000;
    n -= a*500;
    print(res + n/5*5);
}