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;
typedef long long ll;
#define int ll
#define REP(i, n) FOR(i, 0, n)
#define FOR(i, a, b) for (ll i = a; i < b; i++)
#define ALL(a) (a).begin(), (a).end()
typedef vector<ll> vi;
typedef vector<vector<ll>> vvi;
typedef pair<ll, ll> pii;
const ll INF = 1LL << 58;
const ll MOD = 1000000007;
typedef vector<pii> vpii;
template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return true;
    }
    return false;
}
template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return true;
    }
    return false;
}

/* a と b の最大公約数を返す関数 */
long long GCD(long long a, long long b)
{
    if (b == 0)
        return a;
    else
        return GCD(b, a % b);
}

/* a と b の最小公倍数を求める */
long long LCM(long long a, long long b)
{
    long long g = GCD(a, b);
    return a / g * b;
}

long long modinv(long long a, long long m)
{
    long long b = m, u = 1, v = 0;
    while (b)
    {
        long long t = a / b;
        a -= t * b;
        swap(a, b);
        u -= t * v;
        swap(u, v);
    }
    u %= m;
    if (u < 0)
        u += m;
    return u;
}

signed main()
{
    int X;
    cin >> X;
    int ans = 0;
    while (X >= 500)
    {
        X = X - 500;
        ans += 1000;
    }
    while (X >= 5)
    {
        X = X - 5;
        ans += 5;
    }
    cout << ans << endl;
    return 0;
}