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)

/* ----------------------- reclone_rk -------------------------*/ 
#include<bits/stdc++.h>
using namespace std;
#define int long long int
const int mod=1e9+7;
#define re return
#define rep(i, n) for (int i = 0; i < (n); i++)
int gcd(int a, int b){ if(a == 0) return b; return gcd(b%a, a);}
int MOD(int n){ return ((n%mod)+mod)%mod;}
int power(int a,int b){ if(b==0)return 1LL;int s=power(a,b/2LL);s*=s;s%=mod;if(b&1LL)return (s*a)%mod;return s;}
int inv(int a){return power(a,mod-2);}
int nCr(int n,int r){if(r>n)return 0LL;if(n-r<r)r=n-r;int ans=1;for(int i=0;i<r;i++){ans*=(n-i);ans=MOD(ans);ans*=inv(i+1);ans=MOD(ans);}return ans;}

signed main()
{
	int x;
	cin>>x;
	int ans = 0;
	int val = x / 500;
	ans =  1000*val;
	int rem = x%500;
	ans =  ans + (rem)/5*5;

	cout<<ans<<endl;
	re 0;
}