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 PB push_back
#define MP make_pair
#define INS insert
#define LB lower_bound
#define UB upper_bound
#define pii pair <int,int>
#define pll pair <long long, long long>
#define si pair<string, int>
#define is pair<int, string>
#define X first
#define Y second
#define _ << " " <<
#define sz(x) (int)x.size()
#define all(a) (a).begin(),(a).end()
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define FORD(i, a, b) for (int i = (a); i > (b); --i)
#define FORR(i, l, r) for (int i = (l); i <= (r); ++i)
#define FORP(i, a, b) for ((i) = (a); (i) < (b); ++i)
#define FORA(i, x) for (auto &i : x)
#define REP(i, n) FOR(i, 0, n)
#define BITS(x) __builtin_popcount(x)
#define MSET memset
#define MCPY memcpy
#define SQ(a) (a) * (a)
#define TRACE(x) cout << #x " = " << (x) << '\n';
#define YES cout << "YES\n"
#define NO cout << "NO\n"

typedef long long ll;
typedef long double ld;
typedef vector <int> vi;
typedef vector <pii> vpi;
typedef vector <ll> vll;
typedef vector <pll> vpl;
typedef vector <double> vd;
typedef vector <ld> vld;
typedef vector<si> vsi;
typedef vector<is> vis;
typedef vector<string> vs;
//((float) t)/CLOCKS_PER_SEC

const int MOD = 1e9 + 7;
const double PI = acos(-1);
const int INF = 1e9 + 10;
const ll INFL = 1e18 + 10;
const int ABC = 30;
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};
const int dox[] = {-1, 1, 0, 0, -1, -1, 1, 1};
const int doy[] = {0, 0, -1, 1, -1, 1, -1, 1};

inline int sum(int a, int b){
	if (a + b < 0)
		return (a + b + MOD) % MOD;
	return (a + b) % MOD;
}

inline void add(int &a, int b){
	a = sum(a, b);
}

inline int mul(int a, int b){
	return (ll)a * (ll)b % MOD;
}

inline int sub(int a, int b){
	return (a - b + MOD) % MOD;
}

inline int fast_pot(ll pot, int n){
	ll ret = 1;
	while (n){
		if (n & 1)
			ret = (ret * pot) % MOD;
			pot = (pot * pot) % MOD;
			n >>= 1;
		}
	return ret;
}

inline int divide(int a, int b){
	return mul(a, fast_pot(b, MOD - 2));
}

ll lcm(ll a, ll b){
	return abs(a * b) / __gcd(a, b);
}

inline double ccw(pii A, pii B, pii C){
	return (A.X * B.Y) - (A.Y * B.X) + (B.X * C.Y) - (B.Y * C.X) + (C.X * A.Y) - (C.Y * A.X);
}

inline int CCW(pii A, pii B, pii C){
	double val = ccw(A, B, C);
	double eps = max(max(abs(A.X), abs(A.Y)), max(max(abs(B.X), abs(B.Y)), max(abs(C.X), abs(C.Y)))) / 1e9;
	if (val <= -eps)
		return -1;
	if (val >= eps)
		return 1;
	return 0;
}

void to_upper(string &x){
	REP(i, sz(x))
		x[i] = toupper(x[i]);
}

void to_lower(string &x){
	REP(i, sz(x))
		x[i] = tolower(x[i]);
}

string its(ll x){
	if (x == 0)
		return "0";
	string ret = "";
	while (x > 0){
		ret += (x % 10) + '0';
		x /= 10;
	}
	reverse(all(ret));
	return ret;
}

ll sti(string s){
	ll ret = 0;
	REP(i, sz(s)){
		ret *= 10;
		ret += (s[i] - '0');
	}
	return ret;
}

const int N = 2e5 + 10;



int main () {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	ll x; cin >> x;
	cout << (x / 500) * 1000 + ((x % 500) / 5) * 5 << '\n';

	return 0;
}