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.math.BigDecimal;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class C {
    public static void main(String[] args) {
        PrintWriter out = new PrintWriter(System.out);
        InputReader in = new InputReader(System.in);

        int n = in.nextInt();
        int k = in.nextInt();
        char[] s = in.next().toCharArray();

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

        String res;

        if(k > n) {
            StringBuilder builder = new StringBuilder("");
            char min = letters.first();
            for(int i = 0; i < k - n; i++) {
                builder.append(min);
            }
            res = new String(s) + builder;
        } else {
            s = Arrays.copyOf(s, k);
            char max = letters.last();
            for(int i = s.length - 1; i >= 0; i--) {
                if(s[i] == max) {
                    continue;
                } else {
                    boolean flag = false;
                    for(char ch : letters) {
                        if(ch == s[i]) {
                            flag = true;
                        } else if(flag) {
                            s[i] = ch;
                            char min = letters.first();
                            for(int y = i + 1; y < s.length; y++) {
                                s[y] = min;
                            }
                            break;
                        }
                    }
                    break;
                }
            }
            res = new String(s);
        }

        out.print(res);
        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());}
    }
}
