Problem: Welcome to CODE FESTIVAL 2016! In order to celebrate this contest, find a string s that satisfies the following conditions:

* The length of s is between 1 and 5000, inclusive.
* s consists of uppercase letters.
* s contains exactly K occurrences of the string "FESTIVAL" as a subsequence. In other words, there are exactly K tuples of integers (i_0, i_1, ..., i_7) such that 0 ≤ i_0 < i_1 < ... < i_7 ≤ |s|-1 and s[i_0]='F', s[i_1]='E', ..., s[i_7]='L'.



It can be proved that under the given constraints, the solution always exists. In case there are multiple possible solutions, you can output any.

Constraints

* 1 ≤ K ≤ 10^{18}

Input

The input is given from Standard Input in the following format:


K


Output

Print a string that satisfies the conditions.

Examples

Input

7


Output

FESSSSSSSTIVAL


Input

256


Output

FFEESSTTIIVVAALL

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)
#define IREP(i, n) IFOR(i,0,n)
#define Sort(v) sort(v.begin(), v.end())
#define Reverse(v) reverse(v.begin(), v.end())
#define all(v) v.begin(),v.end()
#define SZ(v) ((int)v.size())
#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))
#define Max(a, b) a = max(a, b)
#define Min(a, b) a = min(a, b)
#define bit(n) (1LL<<(n))
#define bit_exist(x, n) ((x >> n) & 1)
#define debug(x) cout << #x << "=" << x << endl;
#define vdebug(v) cout << #v << "=" << endl; REP(i_debug, v.size()){ cout << v[i_debug] << ","; } cout << endl;
#define mdebug(m) cout << #m << "=" << endl; REP(i_debug, m.size()){ REP(j_debug, m[i_debug].size()){ cout << m[i_debug][j_debug] << ","; } cout << endl;}
#define pb push_back
#define f first
#define s second
#define int long long
#define INF 1000000000000000000
template<typename T> istream &operator>>(istream &is, vector<T> &v){ for (auto &x : v) is >> x; return is; }
template<typename T> ostream &operator<<(ostream &os, vector<T> &v){ for(int i = 0; i < v.size(); i++) { cout << v[i]; if(i != v.size() - 1) cout << endl; }; return os; }
template<typename T> void Out(T x) { cout << x << endl; }
template<typename T1, typename T2> void Ans(bool f, T1 y, T2 n) { if(f) Out(y); else Out(n); }

using vec = vector<int>;
using mat = vector<vec>;
using Pii = pair<int, int>;
using PiP = pair<int, Pii>;
using PPi = pair<Pii, int>;
using bools = vector<bool>;
using pairs = vector<Pii>;

//int dx[4] = {1,0,-1,0};
//int dy[4] = {0,1,0,-1};
//char d[4] = {'D','R','U','L'};

const int mod = 1000000007;
//const int mod = 998244353;
#define Add(x, y) x = (x + (y)) % mod
#define Mult(x, y) x = (x * (y)) % mod



signed main(){

    int K; cin >> K;
    const int D = 256;
    vec b(8);
    REP(i, 8){
        b[i] = K % D;
        K /= D;
    }
    string F = "FESTIVAL";

    string ans = "FESTIVA";
    //vec v(8, 1);
    //v[0] = 1;

    REP(d, 8){
        REP(i, b[d]) ans += 'L';
        if(d != 7){
            REP(i, 8){
                IFOR(j, d + 2, 8){
                    //v[j] += v[j - 1];
                    ans += F[j - 1];;
                }
                REP(j, bit(i)) {
                    //v[d + 1] += v[d];
                    ans += F[d];
                }
            }
        }
        //vdebug(v);
    }
    Out(ans);

    return 0;
}