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 Pascal (FPC 2.6.2) (judge verdict: Accepted)

var
	s:String;
	i,j,k:Longint;
	a,b,c,d,e,f:String[5];
begin
	readln(s);
	read(k);
	a:='zzzzz';
	b:=a;
	c:=b;
	d:=c;
	e:=d;
	for i:=1 to length(s)do begin
		for j:=1 to k do begin
			if i+j-1>length(s)then break;
			f:=copy(s,i,j);
			if a>f then begin
				e:=d;
				d:=c;
				c:=b;
				b:=a;
				a:=f;
			end else if a<>f then begin
				if b>f then begin
					e:=d;
					d:=c;
					c:=b;
					b:=f;
				end else if b<>f then begin
					if c>f then begin
						e:=d;
						d:=c;
						c:=f;
					end else if c<>f then begin
						if d>f then begin
							e:=d;
							d:=f;
						end else if d<>f then begin
							if e>f then begin
								e:=f;
							end;
						end;
					end;
				end;
			end;
		end;
	end;
	if k=1 then writeln(a)else if k=2 then writeln(b)else if k=3 then writeln(c)else if k=4 then writeln(d)else writeln(e);
end.