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.

//package com.company;
import java.util.*;
public class Main {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();
        String s = in.next();
        char[] S = s.toCharArray();

        SortedSet<Character> alphabet = new TreeSet();
        for (int i = 0; i < n; i++)
            alphabet.add(S[i]);

        char minChar = alphabet.first();
        if (k > n) {
            System.out.print(s);
            for (int i = 0; i < k - n; i++)
                System.out.print(minChar);
            return;
        }

        if (k < n) {
            char[] newS = new char[k];
            for (int i = 0; i < k; i++)
                newS[i] = S[i];
            S = newS;
            n = S.length;
        }

        Character[] letters = new Character[alphabet.size()];
        letters = alphabet.toArray(new Character[letters.length]);

        int alphaSize = letters.length;

        int lastC = n - 1;

        while (S[lastC] == letters[alphaSize - 1])
            lastC--;

        int index = 0;
        while (letters[index] != S[lastC])
            index++;

        S[lastC] = letters[index + 1];

        for (int i = lastC + 1; i < n; i++)
            S[i] = letters[0];

        System.out.println(S);
    }
}
