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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<string.h>

int main(){
	char s[5001];
	int i,k,one=0,one2=1,g=0,abc[27][5000],abcc[27]={0},j,l;
	scanf("%s%d",s,&k);
	while(s[g]!=0)g++;
	for(i=0;i<g-1;i++){
//		if(s[i-1]>s[i]){
/*
			if(strcmp(&s[one],&s[i])>0){
				one2=one;
				one=i;
			}
			else if(g-k-1>=i&&strcmp(&s[one2],&s[i])>0){
				one2=i;
			}*/
//		}
		if(s[i]!=s[i+1]){
			abc[s[i]-'a'][abcc[s[i]-'a']++]=i;
		}
	}
	for(l=0;l<27;l++){
		for(i=0;i<abcc[l];i++){
			for(j=0;j<abcc[l];j++){
				if(strcmp(&s[abc[l][i]],&s[abc[l][j]])<0){
					int tmp=abc[l][i];
					abc[l][i]=abc[l][j];
					abc[l][j]=tmp;
				}
			}
		}
	}
	/*if(g-one-1<k){
		for(i=0;i<g-one-1;i++){
			putchar(s[one2+i]);
		}
			puts("");
	}else {
		for(i=0;i<k;i++)putchar(s[one+i]);
		puts("");
	}*/
	for(l=0;l<27;l++){
		for(i=0;i<abcc[l];i++){
			if(g-abc[l][i]-1>=k){
				for(j=0;j<k;j++)putchar(s[abc[l][i]+j]);
				puts("");
				return 0;
			}
		}
	}
	return 0;
}