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.util.Arrays;
import java.util.Scanner;
import java.util.TreeSet;

public class C {

    private static int n, k;
    private static char[] s;

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        n = in.nextInt();
        k = in.nextInt();
        s = in.next().toCharArray();

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

        if (k > n) {
            char[] tail = new char[k - n];
            Arrays.fill(tail, chars.iterator().next());
            System.out.println(new String(s) + new String(tail));
            return;
        }

        for (int i = k - 1; i >= 0; i--) {
            if (chars.ceiling((char) (s[i] + 1)) != null) {
                s[i] = chars.ceiling((char) (s[i] + 1));
                for (int j = i + 1; j < k; j++)
                    s[j] = chars.iterator().next();
                break;
            }
        }
        System.out.println(new String(s, 0, k));
    }

}
