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)

#pragma GCC optimize("Ofast") 

////////////////////////////////////////////////////////////////////////////////

#include <bits/stdc++.h>
using namespace std;
#define int long long int
#define endl "\n"
#define debug(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)
#define bew ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define all(x) x.begin(),x.end()
#define pb push_back
#define fi first
#define se second
#define mod (long)(1e9+7)
#define ip(ar,n) for(int i=0;i<n;++i)cin>>ar[i];
#define op(ar,n) for(int i=0;i<n;++i)cout<<ar[i]<<" ";
#define test int t;cin>>t;while(t--)
#define tar(x) int x;cin>>x;
#define fil(ar,val) memset(ar,val,sizeof ar)
#define loop(n) for(int i=0;i<n;++i)

////////////////////////////////////////////////////////////////////////////////

typedef vector<int> vi; 
typedef pair<int, int> ii; 
typedef vector<ii> vii; 
typedef set<int> si;
typedef map<int,int> mi;

////////////////////////////////////////////////////////////////////////////////






signed main() {
bew;
int n;
cin>>n;
int ans=n/500*1000;
n-=n/500*500;
cout<<(ans+n/5*5);




return 0;
}