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 unsigned long long ull;
typedef pair<ll, ll> PA;
typedef priority_queue<int, vector<int>, greater<int> > PQG;
#define REP(i, a, n) for(ll i = (ll) a;i < (ll) n;i++)
#define fi first
#define se second
#define pb push_back
#define INF 1000000009
#define MOD 1000000007
#define MAXN 5003
#define MAXM 200
#define MAXV 1000000009

string s, ans;
int K, L, start, sta;
pair<int, int> len[MAXN];
char mi[6] = {'z','z','z','z','z','z'};

void mikousin(int num){
    REP(i, 0, L){
        len[i] = PA(0, 0);
    }
    mi[num] = 'z';
    REP(i, 0, L){
        if(mi[num] >= s[i] && mi[start] < s[i]){
            mi[num] = s[i];
            if(i < L - 1){
                if(mi[num + 1] >= s[i + 1]){
                    mi[num + 1] = s[i + 1];
                }
            }
        }
    }
    REP(i, 0, L){
        if(s[i] == mi[num]){
            len[i].fi = 1;
            len[i].se = 1;
            if(i == L - 2){
                len[i].fi = 0;
            }
        }
    }
    start++;
}


int main()
{
    cin >> s >> K;
    L = s.size();
    REP(i, 0, L){
        if(mi[0] >= s[i]){
            mi[0] = s[i];
            if(i < L - 1){
                if(mi[1] >= s[i + 1]){
                    mi[1] = s[i + 1];
                }
            }
        }
    }
    REP(i, 0, L){
        if(s[i] == mi[0]){
            len[i].fi = 1;
            len[i].se = 1;
            if(i == L - 2){
                len[i].fi = 0;
            }
        }
    }
    REP(i, 1, K){
        int cnt = 0;
        REP(j, i, L){
            if(len[j - 1].fi == 1 && s[j] == mi[i]){
                len[j].se = len[j - 1].se + 1;
                len[j].fi = 1;
                if(j < L - 1){
                    if(mi[i + 1] >= s[j + 1]){
                        mi[i + 1] = s[j + 1];
                        cnt++;
                    }
                }
                else {
                    len[j].fi = 0;
                }
            }
        }
        if(cnt == 0 && i < K - 1){
            if(i < K - 1){
                mikousin(i + 1);
            }
            sta = i + 1;
        }
    }
    REP(i, sta, K){
        cout << mi[i];
    }
    cout << endl;

    return 0;
}
/*
5 8
0 1 3
1 2 5
2 3 5
3 4 3
4 0 7
0 3 4
2 0 4
4 1 6
*/