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.*;
public class CF940C {
    static char[] ch;
    static int[] count;
    public static void main(String[] args) throws IOException{

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String s[] = br.readLine().split(" ");
        int n = Integer.parseInt(s[0]);
        int k = Integer.parseInt(s[1]);

        count = new int[26];
        String str = br.readLine();
        StringBuffer sb = new StringBuffer(str);
        ch = str.toCharArray();
        for(int i=0;i<n;i++){
            count[(int)ch[i]-97] = 1;
        }

        if(n>=k){
            char smallest = ' ';
            for(int i=0;i<26;i++){
                if(count[i] == 1){
                    smallest = (char)(97+i);
                    break;
                }
            }
            for(int i=k-1;i>=0;i--){
                char c = getNextGreater(ch[i]);
                if(c != ch[i]){
                    ch[i] = c;
                    break;
                }else{
                    ch[i] = smallest;
                }
                //System.out.println(ch[i]);
            }
            sb = new StringBuffer();
            for(int i=0;i<k;i++){
                sb.append(ch[i]+"");
            }
        }else{
            char ch = ' ';
            for(int i=0;i<26;i++){
                if(count[i] == 1){
                    ch = (char)(97+i);
                    break;
                }
            }
            for(int i=0;i<(k-n);i++){
                sb.append(ch+"");
            }
        }

        System.out.println(sb);
    }
    private static char getNextGreater(char ch){
        for(int i=ch-96;i<26;i++){
            if(count[i] == 1){
                return (char)(97+i);
            }
        }
        return ch;
    }
}
