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.util.*;

public class Main {

    public static void main(String[] args) throws Exception {
        new Main().solve();
    }

    private void solve() {
        Scanner scanner=new Scanner(System.in);
        int n=scanner.nextInt(), k=scanner.nextInt();
        char[] chars=scanner.next().toCharArray();

        boolean[] has=new boolean[26];
        char min='z', max='a';
        for (char c: chars) {
            has[c-'a']=true;
            if (min>c) min=c;
            if (max<c) max=c;
        }

        char[] ans=new char[k];

        if (k>n) {
            for (int i=0;i<k;i++) ans[i]=i<n?chars[i]:min;

        }
        else {
            for (int i=k-1;i>=0;i--) {
                if (chars[i]==max) {
                    ans[i]=min;
                }
                else {
                    for (int x=chars[i]+1;;x++) {
                        if (has[x-'a']) {
                            ans[i]=(char)x;
                            break;
                        }
                    }

                    for (int j=i-1;j>=0;j--)
                        ans[j]=chars[j];
                    break;
                }
            }
        }
        System.out.println(new String(ans));
    }
}
