And where the are the phone numbers?

You are given a string s consisting of lowercase English letters and an integer k. Find the lexicographically smallest string t of length k, such that its set of letters is a subset of the set of letters of s and s is lexicographically smaller than t.

It's guaranteed that the answer exists.

Note that the set of letters is a set, not a multiset. For example, the set of letters of abadaba is {a, b, d}.

String p is lexicographically smaller than string q, if p is a prefix of q, is not equal to q or there exists i, such that pi < qi and for all j < i it is satisfied that pj = qj. For example, abc is lexicographically smaller than abcd , abd is lexicographically smaller than abec, afa is not lexicographically smaller than ab and a is not lexicographically smaller than a.

Input

The first line of input contains two space separated integers n and k (1 ≤ n, k ≤ 100 000) — the length of s and the required length of t.

The second line of input contains the string s consisting of n lowercase English letters.

Output

Output the string t conforming to the requirements above.

It's guaranteed that the answer exists.

Examples

Input

3 3
abc


Output

aca


Input

3 2
abc


Output

ac


Input

3 3
ayy


Output

yaa


Input

2 3
ba


Output

baa

Note

In the first example the list of strings t of length 3, such that the set of letters of t is a subset of letters of s is as follows: aaa, aab, aac, aba, abb, abc, aca, acb, .... Among them, those are lexicographically greater than abc: aca, acb, .... Out of those the lexicographically smallest is aca.
import java.io.*;
public class C
{
public static void main(String args[]) throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String in[]=br.readLine().split(" ");
int n=Integer.parseInt(in[0]),k=Integer.parseInt(in[1]);
String s=br.readLine();
int ar[]=new int[26];
int i;
for(i=0;i<s.length();i++)
ar[s.charAt(i)-'a']=1;
int smallest=0,largest=25;
for(i=0;i<26;i++)
if(ar[i]!=0)
{
smallest=i;
break;
}
for(i=25;i>=0;i--)
if(ar[i]!=0)
{
largest=i;
break;
}
if(k>n)
{
System.out.print(s);
for(i=n;i<k;i++)
System.out.print((char)(smallest+'a'));
return;
}
int j,c;
char c1=' ';
for(i=k-1;i>=0;i--)
{
c=s.charAt(i)-'a';
if(c!=largest)
{
for(j=c+1;j<26;j++)
if(ar[j]!=0)
break;
c1=(char)(j+'a');
break;
}
}
//System.out.println(c1);
i=(i>0)?i:0;
System.out.print(s.substring(0,i)+c1);
for(i=i+1;i<k;i++)
System.out.print((char)(smallest+'a'));
}
}