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>
#include<algorithm>
#include<math.h>
#include<string>
#include<iostream>
#include<vector>
#include<cstdlib>
using namespace std;
 
template <class T> using V = vector<T>;
 
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
 
const long long INF = 1LL << 60;
 
using ll = long long;
using db = double;
using st = string;
using ch = char;
using bl = bool;
using vll = V<ll>;
using vpll =V<pair<ll,ll>>;
using vst = V<st>;
using vdb = V<db>;
using vch = V<ch>;
using vbl = V<bl>;
 
 
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define rFOR(i,a,b) for(ll i=(a);i>(b);i--)
#define oFOR(i,a,b) for(ll i=(a);i<(b);i+=2)
#define bgn begin()
#define en end()
#define SORT(a) sort((a).bgn,(a).en)
#define REV(a) reverse((a).bgn,(a).en)
#define M(a,b) max(a,b)
#define rM(a,b) min(a,b)
#define fi first
#define se second
#define sz size()
#define gcd(a,b) __gcd(a,b)
#define co(a) cout<<a<<endl;
#define ci(a) cin>>a;
 

vll dp(100100);

ll sum(ll n) {
ll m=0;
  FOR(i,0,20){
    m+=n%10;
    n/=10;
    if(n==0){
      break;
    }
  }
  return m;
}
 
 
const int MAX = 510000;
const int MOD = 1000000007;
 
long long fac[MAX], finv[MAX], inv[MAX];
 
void Comuse() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
 
 
#define comuse Comuse()
 
ll combi(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
 
 
 
 
ll lcm(ll a,ll b){
  ll n;
  n=a/gcd(a,b)*b;
  return n;
}


 
/****************************************\
| Thank you for viewing my code:)        |
| Written by RedSpica a.k.a. RanseMirage |
| Twitter:@asakaakasaka                  | 
\****************************************/
 
 
signed main() {
  st s;
    ci(s);
  ll k;
    ci(k);
  ll n=s.sz;

  vst A(0);

  FOR(i,0,n){
    FOR(j,1,k+1){
      st t=s.substr(i,j);
      A.push_back(t);
    }
  }

  SORT(A);
  n=A.sz;
  ll count=1;
  FOR(i,1,n){
    if(A[i-1]!=A[i]){
      count++;
    }

    if(count==k){
      co(A[i])
      break;
    }
  }
}