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 ll long long
#define pb push_back
#define vll vector<ll>
#define ub upper_bound
#define lb lower_bound
#define all(v) ((v).begin()),((v).end())
#define ff first
#define ss second
#define mp make_pair
#define pll pair<ll,ll> 
#define fo(i,n) for(ll i=0;i<n;i++)
ll mod1 = 1e9 + 7;
ll mod2 = 998244353;

ll power(ll x,ll y){ll z=1;while(y>0){if(y%2)z=(z*x);x =(x*x) ;y/=2;}return z;}
ll gcd(ll a,ll b){if(a<b) return gcd(b,a);if(b==0) return a;return gcd(b,a%b);} 
ll min(ll a,ll b){if(a<b)return a;return b;}
ll max(ll a,ll b){if(a>b)return a;return b;}







int main(){
	ll n;
	cin>>n;
	ll a = n/500;
	ll b = (n%500)/5;
	cout<<a*1000 + b*5 <<endl;

    
	
}