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)

// https://atcoder.jp/contests/abcXXX/tasks/abcXXX_x

#include <bits/stdc++.h>

using namespace std;
using ll = long long;
using ull = unsigned long long;
using pll = pair<ll, ll>;

static const ll INF = 1001001001;
static const ll LLINF = 1001001001001001001;
static const ll MOD = (ll)1e9 + 7;
static const double EPS = 1e-9;

#define REP(i, n) for(ll i = 0; i < (n); ++i)
#define RANGE(i, m, n) for(ll i = (m); i <= (n); ++i)
#define RRANGE(i, m, n) for(ll i = (m); i >= (n); --i)
#define POSITIVE(x) (x)=((x) < 0) ? 0 : (x)

ll power(ll x, ll y) {
    ll ret = 1;
    while(y-- > 0) ret *= x;
    return ret;
}

ll gcd(ll a, ll b) {
    assert(a >= 0);
    assert(b >= 0);
    if (b == 0) return a;
    return gcd(b, a % b);
}

ll extgcd(ll a, ll b, ll& x, ll& y) {
    x = 1; y = 0;
    assert(a >= 0);
    assert(b >= 0);
    if (b == 0) return a;
    ll X, Y;
    auto d = extgcd(b, a % b, X, Y);
    x = Y;
    y = X - (a / b) * Y;
    return d;
}

int run(istream& in, ostream& out) {
    ll X;
    in >> X;
    ll ans = X/500* 1000;
    X -= X/500*500;
    ans += X/5*5;
    out << ans << endl;
    return 0;
}

int main() {
    return run(cin, cout);
}