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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class C {
    String fileName = "C";

    public void solve() throws IOException {
        int n = nextInt();
        int k = nextInt();
        String s = nextToken();
        char max = s.charAt(0);
        char min = s.charAt(0);
        Set<Character> charsSet = new HashSet<>(26);
        charsSet.add(s.charAt(0));
        for (int i = 1; i < n; i++) {
            char cur = s.charAt(i);
            charsSet.add(cur);
            if (Character.compare(max, cur) < 0) {
                max = cur;
            }
            if (Character.compare(min, cur) > 0) {
                min = cur;
            }
        }
        if (k > n) {
            out.print(s);
            for (int i = 0; i < k - n; i++) {
                out.print(min);
            }
            return;
        }
        List<Character> chars = new ArrayList<>(charsSet);
        int i;
        for (i = k - 1; i >= 0; i--) {
            if (s.charAt(i) != max)
                break;
        }
        int j;
        for (j = 0; j < k - (k - i); j++) {
            out.print(s.charAt(j));
        }
        char nextChar = s.charAt(i);
        out.print(chars.get(chars.indexOf(nextChar) + 1));
        j++;
        for (; j < k; j++) {
            out.print(min);
        }
    }


    public void run() {
        try {
            br = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);

            solve();

            out.close();
        } catch (IOException e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

    BufferedReader br;
    StringTokenizer in;
    PrintWriter out;

    public String nextToken() throws IOException {
        while (in == null || !in.hasMoreTokens()) {
            in = new StringTokenizer(br.readLine());
        }
        return in.nextToken();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    public static void main(String[] args) throws IOException {
        Locale.setDefault(Locale.US);
        new C().run();
    }
}