Problem:
Score: 200 points

Problem Statement

Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it.

The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer).

Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer.

If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.

Constraints

1 \leq N \leq 50000

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

If there are values that could be X, the price of the apple pie before tax, print any one of them.

If there are multiple such values, printing any one of them will be accepted.

If no value could be X, print :(.

Sample Input 1

432

Sample Output 1

400

If the apple pie is priced at 400 yen before tax, you have to pay 400 \times 1.08 = 432 yen to buy one.

Otherwise, the amount you have to pay will not be 432 yen.

Sample Input 2

1079

Sample Output 2

:(

There is no possible price before tax for which you have to pay 1079 yen with tax.

Sample Input 3

1001

Sample Output 3

927

If the apple pie is priced 927 yen before tax, by rounding down 927 \times 1.08 = 1001.16, you have to pay 1001 yen.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define sync ios_base::sync_with_stdio(false), cin.tie(NULL)
#define F first
#define S second
#define pb emplace_back
#define mt make_tuple
#define gcd __gcd
#define PI 3.141592653589
// Input
#define in(a) scanf("%d",&a)
#define in2(a,b) scanf("%d%d",&a,&b)
#define in3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define llin(a) cin >> a
#define inl(a) scanf("%lld",&a)
#define read(v,i,n) for(i=0;i<n;i++)in(v[i])
#define twod(mat,i,j,n,m) rep(i,n){rep(j,m)in(mat[i][j]);}
#define sc(ch) scanf("%c",&ch)
#define sstr(str) scanf("%s",str)
// Output
#define pr(a) printf("%d ",a)
#define pr2(a,b) printf("%d %d\n",a,b)
#define pr3(a,b,c) printf("%d %d %d\n",a,b,c)
#define out(a) printf("%d\n",a)
#define outl(a) printf("%lld\n",a)
#define llpr(a) cout << a << " "
#define llout(a) cout << a << "\n"
#define yes printf("YES\n")
#define no printf("NO\n")
#define lin printf("\n")
// Iterator
#define lp(i,a,b) for(i=a;i<b;i++)
#define rep(i,n) for(i=0;i<n;i++)
#define all(vec) vec.begin(),vec.end()
#define lower(v,k) lower_bound(v.begin(),v.end(),k)-v.begin()
#define upper(v,k) upper_bound(v.begin(),v.end(),k)-v.begin()
#define tf(mytuple) get<0>(mytuple)
#define ts(mytuple) get<1>(mytuple)
#define tt(mytuple) get<2>(mytuple)
// Debug
#define dbg(v,i,n) for(i=0;i<n;i++)pr(v[i]); lin
#define what(x) cerr << #x << " : " << x << "\n"
#define ck printf("continue\n")
#define debug(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args)
{
	cerr << *it << " = " << a << "\n";
	err(++it, args...);
}
// Data Type
#define ll long long int
#define ii pair<int,int>
#define pli pair<ll,int>
#define triple tuple<int,int,int>
#define vi vector<int>
#define vii vector<pair<int,int> >
#define vvi vector<vector<int> >
#define viii vector<pair<pair<int,int>,int> >
#define vvii vector<vector<pair<int,int> > >
#define M 1000000007
#define N 500005
int main()
{
    int i,n,x;
    in(n);
    double d;
    rep(i,n)
    {
        d=i,d*=1.08;
        x=(int)d;
        if(x==n)
        {
            out(i);
            exit(0);
        }
    }
    printf(":(");
    exit(0);
}