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 <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

int len , K , sa[5005] , rk[5005] , he[5005] ;
int tp[5005] , buc[5005] ;
char ss[5005] ;

bool cmp( int *rk , int x , int y , int d ){
	return rk[x] != rk[y] || ( rk[x] == rk[y] && rk[x+d] != rk[y+d] ) ;
}

void Rsort( int *tp , int *rk , int M ){
	for( int i = 0 ; i <= M ; i ++ ) buc[i] = 0 ;
	for( int i = 1 ; i <= len ; i ++ ) buc[ rk[i] ] ++ ;
	for( int i = 1 ; i <= M ; i ++ ) buc[i] += buc[i-1] ;
	for( int i = len ; i ; i -- ) sa[ buc[rk[tp[i]]] -- ] = tp[i] ;
}

void getSA(){
	int *tp = ::tp , *rk = ::rk , m = 127 ;
	for( int i = 1 ; i <= len ; i ++ ) tp[i] = i , rk[i] = ss[i] ;
	Rsort( tp , rk , m ) ;
	for( int l = 1 , p = 0 ; ; l <<= 1 , p = 0 ){
		for( int i = len - l + 1 ; i <= len ; i ++ ) tp[++p] = i ;
		for( int i = 1 ; i <= len ; i ++ )
			if( sa[i] > l ) tp[++p] = sa[i] - l ;
		Rsort( tp , rk , m ) , swap( tp , rk ) , rk[ sa[1] ] = 1 ;
		for( int i = 2 ; i <= len ; i ++ )
			rk[ sa[i] ] = rk[ sa[i-1] ] + cmp( tp , sa[i] , sa[i-1] , l ) ;
		if( rk[ sa[len] ] == len ) break ;
		m = rk[ sa[len] ] ;
	} if( tp != ::tp ) memcpy( rk , tp , sizeof( rk ) ) ;
	for( int i = 1 , j , k = 0 ; i <= len ; he[rk[i]] = k , i ++ )
		for( k=k?k-1:k , j = sa[rk[i]-1] ; ss[i+k] == ss[j+k] ; k ++ ) ;
}

pair<int,int> Query( int K ){
	for( int i = 1 ; i <= len ; i ++ ){
		if( sa[i] - he[i] >= K )
			return make_pair( sa[i] , sa[i] + he[i] + K - 1 ) ;
		else K -= sa[i] - he[i] ;
	}
}

void solve(){
	len = strlen( ss + 1 ) ;
	getSA() ;
	pair<int,int> intval = Query( K ) ;
	for( int i = intval.first ; i <= intval.second ; i ++ )
		printf( "%c" , ss[i] ) ;
}

int main(){
	scanf( "%s%d" , ss + 1 , &K ) ;
	solve() ;
}