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: Runtime Error)

// C
#ifndef _GLIBCXX_NO_ASSERT
#include <cassert>
#endif
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <ciso646>
#include <climits>
#include <clocale>
#include <cmath>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>

#if __cplusplus >= 201103L
#include <ccomplex>
#include <cfenv>
#include <cinttypes>
#include <cstdbool>
#include <cstdint>
#include <ctgmath>
#include <cwchar>
#include <cwctype>
#endif

// C++
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>

#if __cplusplus >= 201103L
#include <array>
#include <atomic>
#include <chrono>
#include <condition_variable>
#include <forward_list>
#include <future>
#include <initializer_list>
#include <mutex>
#include <random>
#include <ratio>
#include <regex>
#include <scoped_allocator>
#include <system_error>
#include <thread>
#include <tuple>
#include <typeindex>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#endif


#define f(i,n) for(int i=0;i<(n);i++)


#define inf (int)(3e18)
#define int long long
#define mod (int)(1000000007)

#define intt long long

using namespace std;

//Library
//モッドパウ
int modpow(int x, int y, int m = mod) {
    int res = 1;
    while (y) {
        if (y % 2) {
            res *= x;
            res %= m;
        }
        x = x * x % mod;
        y /= 2;
    }
    return res;
}

int mypow(int x, int y) {
    int res = 1;
    while (y) {
        if (y % 2) {
            res *= x;
        }
        x = x * x;
        y /= 2;
    }
    return res;
}
//is the number (x) a prime number?
bool prime(int x) {
    for (int i = 2; i * i <= x; i++) {
        if (!(x % i)) {
            return false;
        }
    }
    return true;
}

double kyori(pair<int, int> f, pair<int, int> s) {
    double ans = 0;
    double t = fabs(f.first - s.first);
    double y = fabs(f.second - s.second);
    ans = sqrt(t * t + y * y);
    return ans;
}

//saidai-kouyakusuu
int gcd(int x, int y) {
    if (!y) {
        return x;
    }
    return gcd(y, x % y);
}

//Union-Find Tree

class Union_Find {
    vector<int> par;
    vector<int> rankmy;
public:
    Union_Find(int size) {
        par = vector<int>(size);
        rankmy = vector<int>(size);
        for (int i = 0; i < size; i++) {
            par[i] = i;
        }
    }

    int find(int x) {
        if (par[x] == x) {
            return x;
        }
        return par[x] = find(par[x]);
    }

    void unite(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) {
            return;
        }
        if (rankmy[x] < rankmy[y]) {
            par[x] = y;
        }
        else {
            par[y] = x;
            if (rankmy[x] == rankmy[y]) {
                rankmy[x]++;
            }
        }
    }

    bool same(int x, int y) {
        return find(x) == find(y);
    }

};

//Union-Find-End

//SegTree
template <class T>
class SegTree {
    int n;                       // 葉の数
    vector<T> data;              // データを格納するvector
    T def;                       // 初期値かつ単位元
    function<T(T, T)> operation; // 区間クエリで使う処理
    function<T(T, T)> update;    // 点更新で使う処理

    // 区間[a,b)の総和。ノードk=[l,r)に着目している。
    T _query(int a, int b, int k, int l, int r) {
        if (r <= a || b <= l) return def; // 交差しない
        if (a <= l && r <= b)
            return data[k]; // a,l,r,bの順で完全に含まれる
        else {
            T c1 = _query(a, b, 2 * k + 1, l, (l + r) / 2); // 左の子
            T c2 = _query(a, b, 2 * k + 2, (l + r) / 2, r); // 右の子
            return operation(c1, c2);
        }
    }

public:
    // _n:必要サイズ, _def:初期値かつ単位元, _operation:クエリ関数,
    // _update:更新関数
    SegTree(size_t _n, T _def, function<T(T, T)> _operation,
        function<T(T, T)> _update)
        : def(_def), operation(_operation), update(_update) {
        n = 1;
        while (n < _n) {
            n *= 2;
        }
        data = vector<T>(2 * n - 1, def);
    }

    // 場所i(0-indexed)の値をxで更新
    void change(int i, T x) {
        i += n - 1;
        data[i] = update(data[i], x);
        while (i > 0) {
            i = (i - 1) / 2;
            data[i] = operation(data[i * 2 + 1], data[i * 2 + 2]);
        }
    }

    // [a, b)の区間クエリを実行
    T query(int a, int b) {
        return _query(a, b, 0, 0, n);
    }

    // 添字でアクセス
    T operator[](int i) {
        return data[i + n - 1];
    }
};

#define R_MIN ([](long long a, long long b) { return min(a, b); })
#define R_MAX ([](long long a, long long b) { return max(a, b); })
#define R_SUM ([](long long a, long long b) { return a + b; })

#define NORMAL_UPDATE ([](long long a, long long b) { return b; })
#define ADD_UPDATE ([](long long a, long long b) { return a + b; })
#define MINUS_UPDATE ([](long long a, long long b) { return a - b; }
//Seg-Tree-End

//dfs
vector<int> v[100004];
bool went[100004];
void dfs(int x) {
    went[x] = true;
    for (int i = 0; i < (int)v[x].size(); i++) {
        if (!went[v[x][i]]) {
            dfs(v[x][i]);
        }
    }
}

//number of keta
int keta(int x) {
    int ans = 0;
    while (x) {
        x /= 10;
        ans++;
    }
    return ans;
}

//sum of keta
int ketasum(int x) {
    int ans = 0;
    while (x) {
        ans += x % 10;
        x /= 10;
    }
    return ans;
}

int lcm(int x, int y) {
    int ans = x / gcd(x, y) * y;
    return ans;
}
int twobeki(int x) {
    int ans = 0;
    while (1) {
        if (!(x & 1)) {
            ans++;
            x /= 2;
        }
        else {
            break;
        }
    }
    return ans;
}

int kai(int x, int y) {
    int res = 1;
    for (int i = x - y + 1; i <= x; i++) {
        res *= i; res %= mod;
    }
    return res;
}

int comb(int x, int y) {
    if (y > x)return 0;
    return kai(x, y) * modpow(kai(y, y), mod - 2) % mod;
}

int fukugen(vector<int> l) {
    int ans = 0;
    for (int i = 0; i < (int)l.size(); i++) {
        ans *= 10;
        ans += l[i];
    }
    return ans;
}
#define rep(i,n) for(int i=0;i<n;i++)
//Library-End
int n,k[30],now;
string s;
vector<string> st;
signed main() {
    cin>>s>>n;
    for (int i=0; i<=s.size(); i++) {
        for (int j=1; j<=s.size()-i; j++) {
            st.push_back(s.substr(i,j));
        }
    }
    sort(st.begin(),st.end());
    now=1;
    if (n==1) {
        cout<<st[0]<<endl;
        return 0;
    }
    for (int i=1; i<st.size(); i++) {
        if (st[i-1]!=st[i]) {
            now++;
        }
        if (now==n) {
            cout<<st[i]<<endl;
            return 0;
        }
    }
}