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;
#define MAXN 100000
#define INF 0x3f3f3f3f
#define rint register int
inline int read(){int r;int s=0,c;for(;!isdigit(c=getchar());s=c);for(r=c^48;isdigit(c=getchar());(r*=10)+=c^48);return s^45?r:-r;}
int n, d[MAXN+5], s[MAXN+5], f[MAXN+5][2], g[MAXN+5], h[MAXN+5]; string S = "B", _; vector<int> e[MAXN+5]; int Ans = INF;
void DP(int p, int fa)
{
	s[p] = S[p]=='W'; d[p] = f[p][0] = f[p][1] = g[p] = 0;
	for(auto v:e[p]) if(v^fa) DP(v,p), s[p] += s[v], d[p] += !!f[v][0]; if(!s[p]){f[p][0]=f[p][1]=0; return;}
	h[p] = 0; for(auto v:e[p]) if(v^fa) g[p] += f[v][0], h[p] = min(h[p],f[v][1]-f[v][0]+((d[p]^!!fa^1^(S[p]=='W'))&1));
	f[p][0] = g[p]+((d[p]^!!fa^(S[p]=='W'))&1)+2, f[p][1] = min(f[p][0],g[p]+h[p])+!!fa;
}
int main()
{
	int T = clock(); srand(20010425);
	n = read(); for(rint i = 2, a, b; i <= n; a = read(), b = read(), e[a].push_back(b), e[b].push_back(a), i++);
	cin >> _; S += _; DP(1,0), Ans = f[1][1]; bool flag = s[1];
	do{
		int r; do{r = (rand()<<15|rand())%n+1;}while(flag&&S[r]=='B'); DP(r,0), Ans = min(Ans,f[r][1]);
	}while(clock()-T <= 1700); printf("%d\n",Ans); return 0;
}