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 ss second
#define ff first
#define pb push_back
#define pll pair<long long, long long>
#define FIO ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

ll dx[] = {1,-1,0,0};
ll dy[] = {0,0,1,-1};
const ll sz = 3e5+5,mod = 998244353;
ll t,n,m,k;
vector<ll> p(sz),sp(sz),fac(sz);

const ll lim = 1e18;
const ll inf = 1e15;

ll Power(ll a, ll b)
{
    if(a==0)
        return 0;
    if(b==0)
        return 1;
    ll ans=Power((a*a)%mod,b/2);
    if(b&1)
        return (a*ans)%mod;
    return ans%mod;
}
ll Inv(ll a)
{
    return Power(a,mod-2);
}


int main()
{
    FIO
    cin>>n;
    ll ans = n/500*1000;
    ll x = n%500;
    ans += x/5*5;
    cout<<ans;
}