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.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;

public class Code {
    public static void main(String[] args) {
        // Use the Scanner class
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        sc.nextLine();
        String s = sc.nextLine();
        StringBuilder sb = new StringBuilder();

        char min_char = 'z';
        for (char c : s.toCharArray()) {
            if (c < min_char) {
                min_char = c;
            }
        }

        if (n < k) {
            sb.append(s);
            for (int i = 0; i < k -n; i++) {
                sb.append(min_char);
            }
            System.out.println(sb.toString());
            return;
        }

        Map<Character,Character> m = smallest_character_bigger(s);
        if (n > k) {
            s = s.substring(0,k);
        }
        String res = null;
        for (int j = k-1;j>=0;j--) {
            if (m.containsKey(s.charAt(j))) {
                sb.append(m.get(s.charAt(j)));
                res = s.substring(0,j) + sb.reverse().toString();
                break;
            }
            else {
                sb.append(min_char);
            }
        }
        System.out.println(res);

    }

    static Map<Character,Character> smallest_character_bigger(String s) {
        Map<Character,Character> m = new HashMap<>();
        for (char i = 'a'; i<='z';++i) {
            Character min_char = null;
            for (int j = 0; j < s.length(); ++j) {
                if (s.charAt(j) > i && (min_char == null || s.charAt(j) < min_char)) {
                    min_char = s.charAt(j);
                }
            }
            if (min_char != null) {
                m.put(i, min_char);
            }
        }
        return m;
    }


}