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++ (GCC 9.2.1) (judge verdict: Accepted)

    /**   Created by: Marcos Portales
          Codeforces User: marcosportales
          Country: Cuba
                         **/
#include<bits/stdc++.h>
using namespace std;
/// Macros:
#define int long long
#define f first
#define s second
#define db(x) cerr << #x << ": " << (x) << '\n';
#define pb push_back
#define lb lower_bound
#define up upper_bound
#define all(x) x.begin() , x.end()
#define rall(x) x.rbegin() , x.rend()
#define enl '\n'
#define vi vector<int>
#define sz size
#define rep(i,n) for(int i=0;i<(n);i++)
#define rep2(i,n) for(int i=1;i<=(n);i++)
typedef pair<int,int> ii;
typedef long double ld;
typedef unsigned long long ull;
/// Constraints:
const int maxn = 2*1e5+1;
const int mod = 1000000009;
const ld eps = 1e-9;
const int inf = ((1ll<<31ll)-1ll);
const int INF = 10000000000000000ll;
const ld pi = acos(-1);
/// Prime Numbers:
// 2, 173, 179, 181, 197, 311, 331, 737, 1009, 2011, 2027, 3079, 4001, 10037, 10079, 20011, 20089;
// 100003, 100019, 100043, 200003, 200017, 1000003, 1000033, 1000037, 1000081;
// 2000003, 2000029, 2000039, 1000000007, 1000000021, 2000000099;
/// Functions:
#define lg2(x) 31 - __builtin_clz(x)
#define lgx(x,b) ( log(x) / log(b) )
/// Red-Black Tree Template ---------------------------------
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
//typedef tree < long long ,  null_type ,  less<long long> ,  rb_tree_tag ,  tree_order_statistics_node_update > ordered_set;
/// Quick Pow------------------------------------------------
int qpow(int b,int e){
    if( !e ) return 1;
    if( e & 1 ) return qpow(b,e-1) * b;
    int pwur = qpow(b,e>>1);
    return pwur * pwur;
}
int modinv(int x){
    return qpow(x,mod-2);
}

/// My Code -------------------------------------------------

int32_t main(){
    ios_base::sync_with_stdio(0); cin.tie(0);
    cout.setf(ios::fixed); cout.precision(0);
    srand(time(NULL));

    ///freopen("a.in","r",stdin);
    ///freopen("a.out","w",stdout);

    string s;
    int k;
    cin>>s>>k;
    int n=s.size();
    vector<string>ans(0);
    rep(i,n){
        for(int j=1;(i+j)<=n;j++){
            if(j>k)continue;
            ans.pb(s.substr(i,j));
        }
    }
    sort(all(ans));
    ans.erase(unique(ans.begin(), ans.end()), ans.end());
    cout<<ans[k-1]<<enl;

    return 0;
}