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 sun.reflect.generics.tree.Tree;

import java.io.*;
import java.util.*;

public class C {

    MyScanner in;
    PrintWriter out;

    public static void main(String[] args) throws Exception {
        new C().run();
    }

    public void run() throws Exception {
        in = new MyScanner();
        out = new PrintWriter(System.out);

        solve();

        out.close();
    }

    public void solve() throws Exception {
        int n = in.nextInt();
        int k = in.nextInt();
        char[] s = in.next().toCharArray();

        TreeSet<Character> letters = new TreeSet<Character>();
        for (int i = 0; i < n; i++) {
            letters.add(s[i]);
        }

        if (k <= n) {
            int j = k - 1;
            while (letters.higher(s[j]) == null) {
                j--;
            }

            s[j] = letters.higher(s[j]);
            for (int i = j + 1; i < k; i++) {
                s[i] = letters.first();
            }

            out.println(String.valueOf(s, 0, k));
        } else {
            char[] s2 = new char[k];

            for (int i = 0; i < n; i++) {
                s2[i] = s[i];
            }

            for (int i = n; i < k; i++) {
                s2[i] = letters.first();
            }

            out.println(String.valueOf(s2));
        }
    }

    class MyScanner {

        BufferedReader br;
        StringTokenizer st;

        public MyScanner() throws Exception {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() throws Exception {
            while ((st == null) || (!st.hasMoreTokens())) {
                String t = br.readLine();
                if (t == null) {
                    return null;
                }
                st = new StringTokenizer(t);
            }
            return st.nextToken();
        }

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

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

        boolean nextBoolean() throws Exception {
            return Boolean.parseBoolean(next());
        }

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