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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int, int> P;
map <string,int> mp;
vector<int> V;
#define INF (1<<29)
#define INFLL (1ll<<58)
#define EPS (1e-10)
static const int MAX=1e5;
static const int MOD=1e9+7;
static const int NMAX=50;
static const int MMAX=50;
int dx[4]={0,1,0,-1},dy[4]={-1,0,1,0};
string s;
int k;


void solve(){
    for(int l=1;l<=s.size();l++){//部分文字列の長さ
        for(int i=0;i<s.size();i++){ //i番目からの部分文字列
            if(i+l-1>=s.size())break;
            string ss="";
            for(int j=0;j<l;j++){
                ss+=s[i+j];
            }
            mp[ss]++;
        }
    }
    
    int cunt=0;
    for(auto p: mp){
        if(cunt==k-1) cout<<p.first<<"\n";
        cunt++;
    }
 }

int main(){
    cin.tie(0);  ios::sync_with_stdio(false);
    cin>>s>>k;
    
    solve();

    return 0;
}