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   ipr         	pair<int, int>
#define   pb            push_back
#define   ff            first
#define   ss            second
#define   fr(i, j, k)   for (int i = j; i < k; i++)
#define   rf(i, j, k)   for (int i = j; i >= k;i--)
#define   int           long long
#define   ll 			unsigned long long
#define   ld            long double
#define   all(x)        x.begin(), x.end()
#define   rall(x)       x.rbegin(),x.rend()
#define   mp            make_pair
#define   inf           INT_MAX
#define   read_file();	freopen("input.txt", "r", stdin);	freopen("output.txt", "w", stdout);
#define   IOS           std::ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
 
const int Maxn  =  10000;
const ld  pi   =  acos(-1);
const int mod =  1000000007;
 
 
template<class T> ostream& operator <<(ostream &out,vector<T> &A){for(auto x:A) out<<x;return out;}
template<class T> ostream& operator <<(ostream &out,set<T> &A){for(auto x:A) out<<x<<" ";return out;}
template <class T1,class T2>T1 powr(T1 a,T2 b){T1 res = 1;fr(i,1,b+1)res = res*a;return res;}
int roundUp(int numToRound, int multiple)
{
 if(multiple == 0)
 {
  return numToRound;
 }

 int roundDown = ( (int) (numToRound) / multiple) * multiple;
 int roundUp = roundDown + multiple; 
 int roundCalc = roundUp;
 return (roundCalc);
}
 

 
int32_t main(){IOS
int n;
cin>>n;
cout<<(n/500)*1000 + ((n%500)/5)*5<<endl;

}