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 <iostream>
#include <stdio.h>
#include <algorithm>
#include <vector> 
using namespace std;
int main (){
    char s[5005];
    for(int i=0;i<5005;i++)s[i]='1';
    scanf("%s",s);
    int x;x=1;
    for(int i=0;i<5005;i++){
        if(s[i]=='1')x++;
    }
    int m;
    m=5005-x;
    int k;
    scanf("%d",&k);
    int p;
    p=27*27*27*27*27;
    int a[p];
    for(int i=0;i<p;i++)a[i]=0;
    for(int i=0;i<m;i++){
        a[(s[i]-'a'+1)*p/27]=1;
    }
    for(int i=0;i<m-1;i++){
        a[(s[i]-'a'+1)*p/27+(s[i+1]-'a'+1)*27*27*27]=1;
    }
    for(int i=0;i<m-2;i++){
        a[(s[i]-'a'+1)*p/27+(s[i+1]-'a'+1)*27*27*27+(s[i+2]-'a'+1)*27*27]=1;
    }
    for(int i=0;i<m-3;i++){
        a[(s[i]-'a'+1)*p/27+(s[i+1]-'a'+1)*27*27*27+(s[i+2]-'a'+1)*27*27+(s[i+3]-'a'+1)*27]=1;
    }
    for(int i=0;i<m-4;i++){
        a[(s[i]-'a'+1)*p/27+(s[i+1]-'a'+1)*27*27*27+(s[i+2]-'a'+1)*27*27+(s[i+3]-'a'+1)*27+(s[i+4]-'a'+1)]=1;
    }
    int w;w=1;
    int y;y=0;
    while(w<=k){
        if(a[y]==1)w++;
        y++;
    }
    y--;
    int z;
    int b[5];
    for(int i=0;i<4;i++){
        b[4-i]=y%27;
        z=(y-b[4-i])/27;
        y=z;
    }
    b[0]=y;
    for(int i=0;i<5;i++){
        if(b[i]>0){
            printf("%c",b[i]-1+'a');
        }
    }
printf("\n");
}