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.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;


public class Main {


    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));
        /*
        int n      = sc.nextInt();        // read input as integer
        long k     = sc.nextLong();       // read input as long
        double d   = sc.nextDouble();     // read input as double
        String str = sc.next();           // read input as String
        String s   = sc.nextLine();       // read whole line as String

        int result = 3*n;
        out.println(result);                    // print via PrintWriter
        */
        int n = sc.nextInt();
        int k = sc.nextInt();
        final String s = sc.next();
        Set<Character> set = new HashSet<>();
        for(int i = 0; i < s.length(); i++) {
            set.add(s.charAt(i));
        }
        List<Character> list = new ArrayList<>();
        list.addAll(set);
        Collections.sort(list);
        if (n >= k) {
            char[] res = s.substring(0, k).toCharArray();
            for (int i = res.length - 1; i >= 0; i--) {
                char cur = res[i];
                int index = list.indexOf(cur);
                if (index != list.size() - 1) {
                    res[i] = list.get(index + 1);
                    break;
                } else {
                    res[i] = list.get(0);
                }
            }
            out.println(new String(res));
        } else {
            char[] res = new char[k];
            for (int i = 0; i < n; i++) {
                res[i] = s.charAt(i);
            }
            for (int i = n; i < k; i++) {
                res[i] = list.get(0);
            }
            out.println(new String(res));
        }


        // Stop writing your solution here. -------------------------------------
        out.close();
    }



    //-----------PrintWriter for faster output---------------------------------
    public static PrintWriter out;

    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            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();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

    }
//--------------------------------------------------------
}


