Problem:
Score : 300 points

Problem Statement

You are given a string s.
Among the different substrings of s, print the K-th lexicographically smallest one.

A substring of s is a string obtained by taking out a non-empty contiguous part in s.
For example, if s = ababc, a, bab and ababc are substrings of s, while ac, z and an empty string are not.
Also, we say that substrings are different when they are different as strings.

Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.

Constraints

1 ≤ |s| ≤ 5000

s consists of lowercase English letters.

1 ≤ K ≤ 5

s has at least K different substrings.

Partial Score

200 points will be awarded as a partial score for passing the test set satisfying |s| ≤ 50.

Input

Input is given from Standard Input in the following format:

s
K

Output

Print the K-th lexicographically smallest substring of K.

Sample Input 1

aba
4

Sample Output 1

b

s has five substrings: a, b, ab, ba and aba.
Among them, we should print the fourth smallest one, b.
Note that we do not count a twice.

Sample Input 2

atcoderandatcodeer
5

Sample Output 2

andat

Sample Input 3

z
1

Sample Output 3

z

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

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define pb push_back
#define mp make_pair
#define all(a) begin(a),end(a)
#define FOR(x,val,to) for(int x=(val);x<int((to));++x)
#define FORE(x,val,to) for(auto x=(val);x<=(to);++x)
#define FORR(x,arr) for(auto &x: arr)
#define FORS(x,plus,arr) for(auto x = begin(arr)+(plus); x != end(arr); ++x)
#define FORREV(x,plus,arr) for(auto x = (arr).rbegin()+(plus); x !=(arr).rend(); ++x)
#define REE(s_) {cout<<s_<<'\n';exit(0);}
#define GET(arr) for(auto &i: (arr)) sc(i)
#define whatis(x) cerr << #x << " is " << x << endl;
#define e1 first
#define e2 second
#define INF 0x7f7f7f7f
typedef std::pair<int,int> pi;
typedef std::vector<int> vi;
typedef std::vector<std::string> vs;
typedef int64_t ll;
typedef uint64_t ull;
#define umap unordered_map
#define uset unordered_set
using namespace std;
using namespace __gnu_pbds;

#ifdef _WIN32
#define getchar_unlocked() _getchar_nolock()
#define _CRT_DISABLE_PERFCRIT_LOCKS
#endif
template<class T> ostream& operator<<(ostream &os, vector<T> V) { os<<"[";for(auto const &vv:V)os<<vv<<","; os<<"]"; return os; }
template<class L, class R> ostream& operator<<(ostream &os, pair<L, R> P) { os<<"("<<P.first<<","<<P.second<<")"; return os; }
inline int fstoi(const string &str){auto it=str.begin();bool neg=0;int num=0;if(*it=='-')neg=1;else num=*it-'0';++it;while(it<str.end()) num=num*10+(*it++-'0');if(neg)num*=-1;return num;}
inline void getch(char &x){while(x = getchar_unlocked(), x < 33){;}}
inline void getstr(string &str){str.clear(); char cur;while(cur=getchar_unlocked(),cur<33){;}while(cur>32){str+=cur;cur=getchar_unlocked();}}
template<typename T> inline bool sc(T &num){ bool neg=0; int c; num=0; while(c=getchar_unlocked(),c<33){if(c == EOF) return false;} if(c=='-'){ neg=1; c=getchar_unlocked(); } for(;c>47;c=getchar_unlocked()) num=num*10+c-48; if(neg) num*=-1; return true;}template<typename T, typename ...Args> inline void sc(T &num, Args &...args){ bool neg=0; int c; num=0; while(c=getchar_unlocked(),c<33){;} if(c=='-'){ neg=1; c=getchar_unlocked(); } for(;c>47;c=getchar_unlocked()) num=num*10+c-48; if(neg) num*=-1; sc(args...); }
template<typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; //s.find_by_order(), s.order_of_key() <- works like lower_bound
template<typename T> using ordered_map = tree<T, int, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

int main(){
    ios_base::sync_with_stdio(0);cin.tie(0);
    string s;
    cin >> s;
    int k;
    cin >> k;
    int n = s.size();
    string rt[k+1];
    FORE(i,1,k){
        rt[i] = string(k+2,'z');
    }
    for(int i = 0; i < n;++i){
        string ss;
        for(int x = i; x < min(n, i+k+2); ++x){
            ss += s[x];
            FORE(i,1,k){
                if(ss == rt[i]) break;
                if(ss < rt[i]){
                    for(int j = k-i; j >= i;--j){
                        rt[j+1] = rt[j];
                    }
                    rt[i] = ss;
                    break;
                }
            }
        }
    }
    cout << rt[k] << '\n';
}