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>

using namespace std;

typedef long long ll;
typedef double long dl;
typedef pair < int , int > pii;
typedef pair < pii, pii > p2i;
const int maxn = (int) 3e5 + 1;
const int mod = (int) 1e9 + 7;
const ll inf = (ll)1e18;

#define fi first
#define se second
#define acel ios_base :: sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL)
#define next continue

int n;
string s;
map < string , int > m;

bool cmp(pair < char , int > a, pair < char , int > b){
    while(a.first == b.first){
        a.second++;
        b.second++;
        if(a.second == s.size()){
            return 1;
        }
        if(b.second == s.size()){
            return 0;
        }
        a.first = s[a.second];
        b.first = s[b.second];
    }
    return a.first < b.first;
}

int main(){
    cin >> s >> n;
    for(char c = 'a'; c <= 'z'; c++){
        vector < pair < char , int > > p;
        for(int i = 0; i < (int)s.size(); i++){
            if(s[i] == c){
                if(i + 1 < (int)s.size()){
                    p.push_back({s[i + 1], i + 1});
                }
            }
        }
        if(p.empty()){
            if(s.back() == c){
                n--;
                if(n == 0){
                    cout << c;
                    return 0;
                }
            }
        }
        else {
            sort(p.begin(), p.end(), &cmp);
            for(int i = 0; i < p.size(); i++){
                if((int)s.size() - p[i].second + 1 >= n){
                    cout << c ;
                    n--;
                    for(int j = p[i].second; j < s.size(); j++){
                        if(n <= 0){
                            return 0;
                        }
                        n--;
                        cout << s[j];
                    }
                    return 0;
                }
                else {
                    n -= ((int)s.size() - p[i].second + 2);
                }
            }
            cout << "\n";
        }
    }
}