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.util.*;
import java.io.*;
import java.math.*;
 
public class C
{
    public static void process(int test_number)throws IOException
    {
        int n = ni(), k = ni();
        char s[] = nln().toCharArray();

        if(k > n){
            char take = s[0];
            for(int i = 0; i < n; i++){
                if(Character.compare(take, s[i]) > 0)
                    take = s[i];
                p(s[i]);
            }

            for(int i = 1; i <= k - n; i++)
                p(take);
            p("\n");
        } else{
            TreeSet<Character> set = new TreeSet<Character>();
            ArrayList<Character> li = new ArrayList<>();
            for(char c : s) set.add(c);

            int idx = k - 1;
            for( ; idx >= 0; --idx){
                Character take = set.higher(s[idx]);
                if(take != null){
                    li.add(take);
                    --idx;
                    break;
                } else
                   li.add(set.first());
            }
            
            for( ; idx >= 0; --idx)
                li.add(s[idx]);

            Collections.reverse(li);
            for(char c : li)
                p(c);
            p('\n');
        }
    }
 
    static final long mod = (long)1e9+7l;
    static boolean DEBUG = true;
    static FastReader sc;
    static PrintWriter out;
    public static void main(String[]args)throws IOException
    {
        out = new PrintWriter(System.out);
        sc = new FastReader();
 
        long s = System.currentTimeMillis();
        int t = 1;
        //t = ni();
        for(int i = 1; i <= t; i++)
            process(i);
 
        out.flush();
        System.err.println(System.currentTimeMillis()-s+"ms");
    }

    static void trace(Object... o){ if(!DEBUG) return; System.err.println(Arrays.deepToString(o)); };    
    static void pn(Object o){ out.println(o); }
    static void p(Object o){ out.print(o); }
    static int ni()throws IOException{ return Integer.parseInt(sc.next()); }
    static long nl()throws IOException{ return Long.parseLong(sc.next()); }
    static double nd()throws IOException{ return Double.parseDouble(sc.next()); }
    static String nln()throws IOException{ return sc.nextLine(); }
    static long gcd(long a, long b){ return (b==0)?a:gcd(b,a%b);}
    static int gcd(int a, int b){ return (b==0)?a:gcd(b,a%b); }
    
    static class FastReader{ 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader(){ 
            br = new BufferedReader(new InputStreamReader(System.in)); 
        } 
  
        String next(){ 
            while (st == null || !st.hasMoreElements()){ 
                try{ st = new StringTokenizer(br.readLine()); } catch (IOException  e){ e.printStackTrace(); } 
            } 
            return st.nextToken(); 
        } 
  
        String nextLine(){ 
            String str = ""; 
            try{ str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } 
            return str; 
        } 
    } 
}
