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 Mx 2000001

typedef long long  L;
typedef pair<L, L> pii;
L a,a1,a2,a3,b,b1,b2,b3,c,c1,c2,c3,d,d1,d2,d3,d4,d5,e1,e2,e3,f,h,k,l,m,n,m1,m2,n1,n2,o,p,p1,p2,p3,p4,q,q1,q2,q3,q4,r,s,t,
u,v,x,y,z,sum,mx,mn,flg,flg1,flg2,id1,id2,id3,id4,lim,id;
L A[Mx],B[Mx];
void solve(){
    d1=n/500;
    c=d1*1000;
    n=n%500;
    d1=n/5;
    c+=(d1*5);
    printf("%lld\n",c );
}
int main()
{
    scanf("%lld",&n);
    solve();
    return 0;
}
/// lld I64d push_back() top() pop() insert() upper_bound() lower_bound() memset() max() min() scanf continue