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.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        StringBuilder str = new StringBuilder(sc.next());
        byte[] alph = new byte['z' - 'a' + 1];

        char min = 'z';
        char max = 'a';
        for (int i = 0; i < str.length(); i++) {
            int index = str.charAt(i) - 'a';
            alph[index] = 1;
            if (str.charAt(i) > max) max = str.charAt(i);
            if (str.charAt(i) < min) min = str.charAt(i);
        }

        if (k > n) {
            for (int i = 0; i < k - n; i++) {
                str.append(min);
            }
            System.out.println(str);
        } else {
            StringBuilder sub = new StringBuilder(str.substring(0, k));
            for (int i = k - 1; i >= 0; i--) {
                char last = sub.charAt(i);
                if (last < max) {
                    int index = last - 'a' + 1;
                    for (int j = index; j < 32; j++) {
                        if (alph[j] == 1) {
                            index = j;
                            break;
                        }
                    }
                    sub.setCharAt(i, (char) (index + 'a'));
                    for (int j = i + 1; j < k; j++) {
                        sub.setCharAt(j, min);
                    }
                    break;
                }
            }
            System.out.println(sub);
        }
    }
}
