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.*;
import java.util.*;
 
public class Main {
  
  public static void main(String[] args)throws IOException {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int n = in.nextInt();
        int k = in.nextInt();
        String s = in.next();
        int ix = -1;
        char arr[] = new char[k];
        int alpha[] = new int[26];
        for(int i=0;i<26;i++)
          alpha[i]=0;
        for(int i=0;i<n;i++)
          alpha[s.charAt(i)-'a']++;
        for(int i=0;i<k;i++)
        {
          if(i>=n)
          {
            ix=i;
            continue;
          }
          char c = s.charAt(i);
          boolean exists=false;
          for(int j=(c-'a')+1;j<26;j++)
            if(alpha[j]>0)
              exists=true;
          if(exists)
            ix=i;
        }
        // out.println("ix="+ix);
        char minchar = 'a';
        for(int i=0;i<26;i++)
          if(alpha[i]>0){
            minchar=(char)('a'+i); break;
          }
        char c=(char)'a'-1;
        if(ix<n)
          c=s.charAt(ix);
        for(int i=0;i<26;i++)
            if(alpha[i]>0 && (char)('a'+i)>c)
            {
              c=(char)('a'+i);
              break;
            }
        for(int i=0;i<ix;i++)
        {
          if(i<n)
          {
            out.print(s.charAt(i));
          }
          else
          {
            out.print(minchar);
          }
        }
        out.print(c);
        for(int i=ix+1;i<k;i++)
        {
          out.print(minchar);
        }
        out.println();
        out.close();
  }
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
 
        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }
 
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
 
        public int nextInt() {
            return Integer.parseInt(next());
        }
 
        public long nextLong() {
            return Long.parseLong(next());
        }
 
    }
 
}