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: Accepted)

#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
 
struct trie{
    int next[26],cnt;
    bool flag;
    trie():flag(false){
    	cnt=0;
        for (int i=0; i<26; i++)
            next[i]=0;
    }
} tr[1000005];
int n,m,root,ip,len,k;
char ch[5005];
string ans;
 
inline void build(int now)
{
    len=strlen(ch);
    int cur=root;
    int id;
    int cen=0;
    for (int i=now-1; i<len; i++)
    {
        if (cen>5) break;
        id=ch[i]-'a';
        if (tr[cur].next[id]==0) {
            tr[cur].next[id]=++ip;
        }
        cur=tr[cur].next[id];
        cen++;
    }
    tr[cur].flag=true;
}
 
void dfs(int x)
{
    int cur=x;
    tr[cur].cnt=1;
    for (int i=0; i<26; i++)
    {
        if (tr[cur].next[i])
        {
            dfs(tr[cur].next[i]);
            tr[cur].cnt+=tr[tr[cur].next[i]].cnt;
        }
    }
}
 
inline void find(int kth)
{
    int cur=root;
    while (kth)
    {
        for (int i=0; i<26; i++)
        { 
            if (tr[cur].next[i])
            {
                if (tr[tr[cur].next[i]].cnt>=kth)
                {
                    cur=tr[cur].next[i];
                    kth--;
                    ans+=i+'a';
                    break;
                }
                else {
                    kth-=tr[tr[cur].next[i]].cnt;
                }
            }
        }
    }
}
 
int main()
{
    scanf("%s",ch);
    n=strlen(ch);
    for (int i=1; i<=n; i++)
        build(i);
    dfs(root);
         
    scanf("%d",&k);
    find(k);
     
    cout<<ans;
     
    return 0;
}