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.io.*;
import java.lang.*;
import java.util.*;

public class C940 {
    public static void main(String[] args) throws IOException{
        StringBuffer ans = new StringBuffer();
        StringTokenizer st;
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        st = new StringTokenizer(f.readLine());
        int n = Integer.parseInt(st.nextToken());
        int k = Integer.parseInt(st.nextToken());
        st = new StringTokenizer(f.readLine());
        String str = st.nextToken();
        f.close();
        ArrayList<Integer> arrl = new ArrayList<>();
        HashSet<Integer> hash = new HashSet<>();
        for(int i = 0; i < n; i++){
            if(!hash.contains((int)str.charAt(i))){
                hash.add((int)str.charAt(i));
                arrl.add((int)str.charAt(i));
            }
        }
        Collections.sort(arrl);
        if(k > n)
            ans.append(str);
        else if(k < n){
            int d = n-k;
            int cnt = 0;
            while (str.charAt(n-d-1) == arrl.get(arrl.size()-1)){
                d++;
                cnt++;
            }
            d = k - cnt - 1;
            //System.out.println(d);
            //System.out.println(d);
            ans.append(str, 0, d);
            int r = Collections.binarySearch(arrl, (int)str.charAt(d));
            ans.append(Character.toString(arrl.get(r+1)));
            while (ans.length() < k)
                ans.append(Character.toString(arrl.get(0)));
        }else{
            int d = 0;
            while (str.charAt(n-d-1) == arrl.get(arrl.size()-1)){
                d++;
            }
            d = n-d-1;
            ans.append(str, 0, d);
            int r = Collections.binarySearch(arrl, (int)str.charAt(d));
            ans.append(Character.toString(arrl.get(r+1)));
            while (ans.length() < k)
                ans.append(Character.toString(arrl.get(0)));
        }
        while (k > str.length() && k > ans.length()){
            ans.append(Character.toString(arrl.get(0)));
        }




        System.out.println(ans);
    }

}