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 sun.reflect.generics.tree.Tree;

import java.io.PrintStream;
import java.util.*;

public class Main {

    public static void main(String[] args) throws Exception {
//        System.setOut(new PrintStream("output.txt"));
        Scanner scanner=new Scanner(System.in);
//        int times=scanner.nextInt();
        int times=1;
        long start=System.currentTimeMillis();
        for (int t=1;t<=times;t++) {
//            System.out.println(String.format("Case #%d: %s", t, new Main().solve(scanner)));
            System.out.println(new Main().solve(scanner));
        }
        long end=System.currentTimeMillis();
        System.err.println(String.format("Time used: %.3fs", (end-start)/1000.0));
    }

    public String solve(Scanner sc) {
        int n=sc.nextInt(), k=sc.nextInt();
        String s=sc.next();
        StringBuilder sb = new StringBuilder();
        TreeSet<Character> set = new TreeSet<>();
        for(int i=0; i<n&&set.size()<26; i++){
            set.add(s.charAt(i));
        }

        if(k>n) {
            sb.append(s);
            while(n<k){
                sb.append(set.first()); n++;
            }
            return sb.toString();
        }

        for(int i=k-1; i>=0; i--){
            if(s.charAt(i)!=set.last()) {
                sb.insert(0, set.higher(s.charAt(i)));
                return s.substring(0, i)+sb.toString();
            }else sb.append(set.first());
        }
        return sb.toString();
    }


}
