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: Accepted)

#include <bits/stdc++.h>
using namespace std;

template <class T> using V = vector<T>;
template <class S, class T> using P = pair<S, T>;

using ll = long long;
using ull = unsigned long long;
using vll = V<ll>;
using vvll = V<vll>;
using vvvll = V<vvll>;
using pl = P<ll, ll>;
using vpl = V<pl>;
using vvpl = V<vpl>;
using vs = V<string>;
using qll = queue<ll>;
using qpl = queue<pl>;
using mapll = map<ll, ll>;
using setll = set<ll>;
using pqll = priority_queue<ll>;

#define int ll
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pob pop_back()
#define pf push_front
#define pof pop_front()
#define sz size()
#define bgn begin()
#define en end()
#define emp empty()
#define fr front()
#define bk back()
#define res resize
#define tp top()
#define p_q priority_queue
#define inv inverse()

#define FOR(i,a,b) for(ll i=(a);i<=(ll)(b);i++)
#define rFOR(i,a,b) for(ll i=(b);i>=(ll)(a);i--)
#define REP(i,a) FOR((i),0,(ll)(a)-1)
#define REP0(i,a) FOR((i),0,(ll)(a))
#define REP1(i,a) FOR((i),1,(ll)(a))
#define rREP(i,a) rFOR((i),0,(ll)(a)-1)
#define rREP0(i,a) rFOR((i),0,(ll)(a))
#define rREP1(i,a) rFOR((i),1,(ll)(a))
#define IOTA(a,n) iota((a).bgn,(a).en,(n))
#define SORT(a) sort((a).bgn,(a).en)
#define rSORT(a) sort((a).rbegin(),(a).rend())
#define UNIQUE(a) (a).erase(unique((a).bgn,(a).en),(a).en)
#define BINS(a,b) binary_search((a).bgn,(a).en,(b))
#define LOWB(a,b) (lower_bound((a).bgn,(a).en,(b))-(a).bgn)
#define UPB(a,b) (upper_bound((a).bgn,(a).en,(b))-(a).bgn)
#define CNT(a,b) count((a).bgn,(a).en,b)
#define SUM(a) accumulate((a).bgn,(a).en,0)
#define REV(a) reverse((a).bgn,(a).en)
#define yn(a) cout <<((a)?"yes":"no")<<"\n";
#define Yn(a) cout <<((a)?"Yes":"No")<<"\n";
#define YN(a) cout <<((a)?"YES":"NO")<<"\n";
#define imp(a) cout <<((a)?"possible":"impossible")<<"\n";
#define Imp(a) cout <<((a)?"Possible":"Impossible")<<"\n";
#define IMP(a) cout <<((a)?"POSSIBLE":"IMPOSSIBLE")<<"\n";
#define dbg(a) cerr <<(#a)<<": "<<(a)<<"\n";
#define DigN2(a) ((llabs(a)==0)?(1):((ll)(log2(double(llabs(a))))+1))
#define DigN10(a) ((llabs(a)==0)?(1):((ll)(log10(double(llabs(a))))+1))
#define Dig2(a,b) (((a)>>(b))&1)
#define Dig10(a,b) (ll)(((a)/((ll)(pow(10.0,(double)(b)))))%10)
#define Pow2(a) ((ll)(1)<<(a))
#define Pow10(a) ((ll)(pow(10.0,double(a))))
#define powll(a,b) (ll)(pow((double)(a),(double)(b)))

#define li(...) ll __VA_ARGS__;Input(__VA_ARGS__);
#define si(...) string __VA_ARGS__;Input(__VA_ARGS__);
#define vli(size, ...) vll __VA_ARGS__;vInput(size,__VA_ARGS__);
#define vsi(size, ...) vs __VA_ARGS__;vInput(size,__VA_ARGS__);

const ll MOD = 1e9 + 7;
//const ll MOD = 998244353;
//const ll MOD = 924844033;
//const ll MOD = 9007199254740881;
const ll INF = 1LL << 60;//1.15e18
const double PI = acos(-1.0);
const string alp = "abcdefghijklmnopqrstuvwxyz";
const string ALP = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

void Input() {}
template <class Var, class... Args> void Input(Var& var, Args&... args) {
    cin >> var;
    Input(args...);
}

void vInit(ll size) {}
template <class T, class... Args> void vInit(ll size, V<T>& v, Args&... args) {
    v.res(size);
    vInit(size, args...);
}
void vInputNum(ll num) {}
template <class T, class... Args> void vInputNum(ll num, V<T>& v, Args&... args) {
    cin >> v[num];
    vInputNum(num, args...);
}
void vInput(ll size) {}
template <class... Args> void vInput(ll size, Args&... args) {
    vInit(size, args...);
    REP(i, size) vInputNum(i, args...);
}

template <class S, class T> ostream &operator<<(ostream &out, const P<S, T> &p) {
    return out << "[" << p.fi << ", " << p.se << "]";
}
 
template <class T> ostream &operator<<(ostream &out, V<T> &v) {
    if (v.emp) return out << "{}";
    else {
        auto itr = v.bgn;
        out << "{" << *itr;
        itr++;
        while (itr != v.en) {
            out << ", " << *itr;
            itr++;
        }
        out << "}";
        return out;
    }
}
 
template <class S, class T> ostream &operator<<(ostream &out, const map<S, T> &m) {
    if (m.emp) return out << "<[]>";
    else {
        auto itr = m.bgn;
        out << "< [" << (itr->fi) << ": " << (itr->se);
        itr++;
        while (itr != m.en) {
            out << "], [" << (itr->fi) << ": " << (itr->se);
            itr++;
        }
        out << "] >";
        return out;
    }
}
 
template <class T> ostream &operator<<(ostream &out, const set<T> &s) {
    if (s.emp) return out << "<>";
    else {
        auto itr = s.bgn;
        out << "<" << *itr;
        itr++;
        while (itr != s.en) {
            out << ", " << *itr;
            itr++;
        }
        out << ">";
        return out;
    }
}

ll gcd(ll a, ll b) { if (a < b) { a ^= b; b ^= a; a ^= b; } return b ? gcd(b, a%b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b)*b; }

struct UFT {

public:
    ll tsize;
    ll mode;
    vll par;
    vll rank;
    UFT() {}
    UFT(const UFT &uft) {}
    UFT(ll tsizeget, ll modeget = 0){
        tsize = tsizeget;
        mode = modeget;
        par.assign(tsize, -1);
        if (!mode) rank.res(tsize, 0);
    }
    ll root(ll x) {
        return par[x] < 0 ? x : par[x] = root(par[x]);
    }
    bool isRoot(ll x) {
        return x == root(x);
    }
    bool same(ll x, ll y) {
        return root(x) == root(y);
    }
    void merge(ll x, ll y) {
        x = root(x);
        y = root(y);
        if (x == y) return;
        if (mode) {
            par[x] += par[y];
            par[y] = x;
        }
        else {
            if (rank[x] < rank[y]) {
                par[y] += par[x];
                par[x] = y;
            }
            else {
                par[x] += par[y];
                par[y] = x;
                if (rank[x] == rank[y]) rank[x]++;
            }
        }
    }
    ll size(ll x) {
        return -par[root(x)];
    }
};

ll isP(ll n) {
    if (n <= 1) return 0;
    FOR(i, 2, (ll)sqrt(n) + 1) {
        if (n%i == 0) return 0;
    }
    return 1;
}

vvll CombMemo(3000, vll(3000, -1));

ll Comb(ll n, ll k) {
    if ((n < 0) || (k < 0)) return 0;
    if (CombMemo[n][k] == -1) {
        if (n < k) CombMemo[n][k] = 0;
        else {
            if (n == 0) CombMemo[n][k] = 1;
            else if (k == 0) CombMemo[n][k] = 1;
            else if (n == k) CombMemo[n][k] = 1;
            else CombMemo[n][k] = Comb(n - 1, k - 1) + Comb(n - 1, k);
        }
    }
    return CombMemo[n][k];
}

void Solve();

signed main() { 
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << setprecision(20) << fixed;
    Solve();
}

void Solve() {
  li(n);
  REP1(i,50000){
    if ((ll)(i*1.08)==n) {
      cout << i;
      return;
    }
  }
  cout << ":(";
}