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.math.BigInteger;
import java.util.Arrays;
import java.util.Comparator;
import java.util.Random;
import java.util.StringTokenizer;

public class Main
{

    public Main() throws IOException
    {

    }

    private void solve() throws IOException
    {
        int n = nextInt();
        int k = nextInt();
        String s = next();
        if (k > n)
        {
            int mx = 26;
            for (int i = 0; i < n; i++)
            {
                mx = Math.min(mx, s.charAt(i) - 'a');
            }
            out.print(s);
            for (int i = 0; i < k - n; i++) out.print((char)(mx + 'a'));
            out.println();
        }
        else
        {
            int mx = 0;
            for (int i = 0; i < n; i++)
            {
                mx = Math.max(mx, s.charAt(i) - 'a');
            }
            int mn = 26;
            int go = -1;
            for (int i = k - 1; i >= 0; i--)
            {
                if (s.charAt(i) != (char) (mx + 'a'))
                {
                    go = i;
                    break;
                }
            }
            for (int i = 0; i < n; i++)
            {
                if (s.charAt(i) > s.charAt(go))
                {
                    mn = Math.min(mn, s.charAt(i) - 'a');
                }
            }
            for (int i = 0; i < go; i++) out.print(s.charAt(i));
            out.print((char) (mn + 'a'));
            for (int i = 0; i < n; i++)
            {
                mn = Math.min(mn, s.charAt(i) - 'a');
            }
            for (int i = 0; i < k - go - 1; i++) out.print((char) (mn + 'a'));
            out.println();
        }
    }

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

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st;
    PrintWriter out = new PrintWriter(System.out);
    String next() throws IOException
    {
        while (st == null || !st.hasMoreTokens())
        {
            st = new StringTokenizer(br.readLine());
        }
        return st.nextToken();
    }

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

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

    private void run() throws IOException
    {
        solve();
        out.close();
    }

}