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)

#include<cstdio>
#include<cstring>

using namespace std;

char s[5005];
int len, K, sum;

struct Trie{
	int c[26], pre;
	char ch;
}tr[5005];
int cnt;
void insert(int pos){
	int now = 0;
	for(int i = pos; i <= len; i++){
		if(tr[now].c[s[i]-'a'])
			now = tr[now].c[s[i]-'a'];
		else{
			tr[now].c[s[i]-'a'] = ++cnt;
			tr[cnt].pre = now;
			tr[cnt].ch = s[i];
			now = cnt;
		}
	}
}

char ans[5005];
void print(int x){
	while(x){
		ans[++ans[0]] = tr[x].ch;
		x = tr[x].pre;
	}
	for(int i = ans[0]; i; i--)
		putchar(ans[i]);
}

bool findAns;
void dfs(int x){
	if(findAns)	return;
	if(sum == K){
		findAns = 2;
		print(x);
		return;
	}
	for(int i = 'a'; i <= 'z'; i++)
		if(tr[x].c[i-'a'])
			sum++, dfs(tr[x].c[i-'a']);
}

int main(){
	scanf("%s", s+1);
	scanf("%d", &K);
	len = strlen(s+1);
	for(int i = 1; i <= len; i++)
		insert(i);
	dfs(0);
	return 0;
}