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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;


public class vk18
{
    public static void main(String[] stp) throws Exception
    {
        Scanner scan = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken()), i,k=Integer.parseInt(st.nextToken());
        st=new StringTokenizer(br.readLine());
        char []c=st.nextToken().toCharArray();
        Integer a[]=new Integer[26];
        Arrays.fill(a,0);
        for(i=0;i<c.length;i++) a[(int)(c[i]-'a')]++;
        char []ans=new char[k];
        char min='P';
        for(i=0;i<26;i++)
        {
            if(a[i]!=0) {min=(char)(i+'a'); break; }
        }
        if(n < k)
        {

            for(i=0;i<n;i++) ans[i]=c[i];
            for(i=n;i<k;i++) ans[i]=min;
        }
        else
        {
            for(i=0;i<k;i++) ans[i]=c[i];
            boolean flag=false;
            for(i=k-1;i>=0;i--)
            {
                int j=(int)(ans[i]-'a');
                for(j=j+1;j<26;j++)
                {
                    if(a[j]!=0) { ans[i]=(char)(j+'a'); flag=true; break;}
                }
                if(flag) break;
                else
                {
                    ans[i]=min;
                }
            }
        }
        pw.println(String.valueOf(ans));
        pw.flush();
    }
}