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.*;

public class PhoneNumbers {
    public static void main(String[] args){
        Scanner s = new Scanner(System.in);
        s.nextInt();
        int k = s.nextInt();
        String tmp = s.nextLine();
        String str = s.nextLine();
        System.out.println(PhoneNumbers.solve(k, str));
    }

    public static String solve(int k, String s) {
        int n = s.length();
        Set<Character> abc = new HashSet<>();
        for (char ch : s.toCharArray()) {
            abc.add(ch);
        }
        char[] abcOrd = new char[abc.size()];
        int abcLen = abc.size();
        Map<Character, Integer> map = new HashMap<>();
        int i = 0;
        for (char ch : abc) {
            abcOrd[i++] = ch;
        }
        Arrays.sort(abcOrd);
        for(i = 0; i<abcOrd.length; i++) {
            map.put(abcOrd[i], i);
        }

        char[] result = new char[k];
        for (int j = 0; j < k; j++) {
            if(j<n) {
                result[j] = s.charAt(j);
            } else {
                result[j] = abcOrd[0];
            }
        }
        if(k>n) return new String(result);

        int p = k-1;
        while (p >= 0) {
            int idx = map.get(result[p]);
            if (idx == (abcLen - 1)) {
                result[p--] = abcOrd[0];
            } else {
                result[p] = abcOrd[idx + 1];
                break;
            }
        }

        return new String(result);
    }
}
