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>
#include <ext/pb_ds/detail/standard_policies.hpp>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp>
#define ll long long
#define int long long 
#define pb push_back
#define ins insert
#define rs resize
#define br break;
#define cont continue;
#define vi vector<int>
#define vll vector<long long>
#define si set<ll>
#define sll set<long long>
#define pii pair<ll,ll>
#define pll pair<long long, long long>
#define speed ios_base::sync_with_stdio(false);cin.tie(NULL);
#define vec vector
#define cend cout<<endl;
#define sc second
#define fr first
#define retz return 0;
#define ll_MAX (ll)(1e9*1e9)
#define ll_MIN (ll)(-1e9*1e9)
#define lb lower_bound
#define ub upper_bound
#define rep(i,a,b)  for(int i=a; i<=b; i++)
#define repd(i,a,b) for(int i=a; i>=b; i--)
#define ps(arr, n) for(int i=1; i<n; i++)arr[i]+=arr[i-1];
#define inp(arr, n)for(int i=0; i<n; i++)cin>>arr[i];
#define out(arr, n)for(int i=0; i<n; i++)cout<<arr[i];
#define inp1(arr,n,m)for(int i=0; i<n; i++)for(int j=0;j<m;j++)cin>>arr[i][j];
#define out1(arr,n,m)for(int i=0; i<n; i++)for(int j=0;j<m;j++)cin>>arr[i][j];
#define loop(q) for(auto it:q)
#define loop1(q) for(auto&it:q)
#define db1 cout << "*" << endl
#define db2 cout << "YES" << endl
using namespace std;
using namespace __gnu_pbds; 
#define oset(X) tree<X,null_type,less<X>, rb_tree_tag, tree_order_statistics_node_update> 
#define roset(X) tree<X,null_type,greater<X>, rb_tree_tag, tree_order_statistics_node_update>  
#define okey order_of_key //lower_bound_set
#define ofind find_by_order
#define cyes cout << "Yes" << endl; 
#define cno cout << "No" << endl;
#define mod (long long)1000000007
#define mod2 998244353
#define INT_MAX1 900000000
#define MAXN 2000000
#define INF INT_MAX





void solve()
{
  int n;
  cin >> n;
  int x = n/500;
  n%=500;
  x = x*1000+(n/5)*5;
  cout << x<< endl;
}

signed main()
{
    
    speed;
    int t = 1;
    //cin >> t;
    while(t--)
        solve();
}