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: Accepted)

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


void copy(char *to,char *from,int n){
	int i;
	for(i=0;i<6;i++){
		if(i<n)	to[i] = from[i];
		else to[i] = '\0';
	}
}


int check(char *a,char *b){
	int i;
	for(i=0;i<5;i++){
		if(a[i] < b[i])			return -1;	//a -> b
		else if (a[i] > b[i])	return 1;	//b -> a
	}
	return 0;	//a == b
}

int int_sort( const void * a , const void * b ) {
  return strcmp(( char * )a , ( char * )b );
}

int main(void){
	char s[5001];
	int K;
	int i,j;
	scanf("%s",s);
	scanf("%d",&K);
	
	char kouho[25000][6];
	int index = 0;
	
	
	for(i=0;s[i] != '\0';i++){
		for(j=1; s[i+j-1] != '\0' ;j++){
			if(j>5) break;
			copy(kouho[index],&s[i],j);
			index++;
		}
	}
		
	
	qsort(kouho,index,sizeof(kouho[0]),int_sort);

	int ex[25000];
	for(i=0;i<25000;i++){
		ex[i] = 1;
	}
	for(i=0;i<index;i++){
		if(check(kouho[i],kouho[i+1]) == 0){
			ex[i+1] = -1;
		}
	}
	
		
	int count = 0,flag = 0;
	i = 0;
	while(flag <= K){
		if(ex[i] > 0){
			count = i;
			flag++;
		}
		i++;
	}
	
	
	printf("%s",kouho[count-1]);
	
	
	return 0;}