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.
//package  com.company;
import java.io.*;
import java.util.*;
public class Main{
    public static void main(String[] args){
        Scanner s=new Scanner(System.in);
       int n=s.nextInt();
       int  k=s.nextInt();
       char[] a=s.next().toCharArray();
       char[] b=a.clone();
       Arrays.sort(b);int flag=0;String ans="";
       char pre=b[0];
       HashMap<Character,Character> h=new HashMap<Character, Character>();
       for(int i=1;i<n;i++){
       if(b[i]!=pre){
           h.put(pre,b[i]);pre=b[i];
       }
       }if(k>n){for(int i=0;i<n;i++)
            System.out.print(a[i]);
       for(int i=n;i<k;i++) System.out.print(b[0]);
        }else{char[] v=new char[k];
           for(int i=k-1;i>=0;i--){
               if(flag==0&&a[i]!=b[n-1]){
                   flag=1; v[i]=h.get(a[i]);
               }else if(flag==0&&a[i]==b[n-1]){
                   v[i]=b[0];
               }else if(flag==1){
                   v[i]=a[i];
               }
           }for(int i=0;i<k;i++){
                System.out.print(v[i]);
                }
            }
        }













}