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;
typedef long double ld;
typedef pair<ll, ll> l_l;
typedef pair<int , int> i_i;
typedef vector<ll> vl;
typedef vector<int> vi;
typedef vector<char> vec;
typedef vector<bool> veb;
typedef vector<string> ves;
typedef vector<vector<ll>> vvl;
typedef vector<vector<int>> vvi;
typedef vector<vector<char>> vvec;
typedef vector<vector<bool>> vveb;
typedef vector<vector<string>> vves;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<(int)(n);i++)
#define rep2(i,n) for(int i=2;i<(int)(n);i++)
#define repk(i,k,n) for(int i=k;i<(int)(n);i++)
#define rev(v) reverse(v.begin(), v.end())
#define fs first
#define sc second
#define pb push_back
#define pp pop_back
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define maxel(a) *max_element(all(a))
#define minel(a) *min_element(all(a))
#define acc accumulate
#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
#define mod (1000000007)
typedef long long int64;
//const int64 INF = 1LL << 58;
#define dame { puts("-1"); return 0;}
#define YES { cout << "YES" << endl; return 0; }
#define NO { cout << "NO" << endl; return 0; }
#define Yes { cout << "Yes" << endl; return 0; }
#define No { cout << "No" << endl; return 0; }
#define uniq(a) sort(all(a));(a).resize(unique(all(a)) - a.begin());
#define dab(v) for(auto i:v)cout<<i<<' '; cout << endl;
#define vin(v) for (ll i = 0; i < v.size(); ++i) cin >> v[i];
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
int dx[] = {0, 0, -1, 1};
int dy[] = {1, -1, 0, 0};

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
    //cout << fixed << setprecision(15);
    
    ll X; cin>>X;
	int ans = 0;
	ans += (X/500) * 1000;
	X %= 500;
	ans += X / 5 * 5;
	cout << ans << endl;
}