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 P940C
{
    public static void main(String[] args)
    {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();
        char[] s = in.next().toCharArray();
        boolean[] appeared = new boolean[256];
        for(int i = 0; i < n; ++i)
        {
            appeared[s[i]] = true;
        }

        char m = 'a';
        while(m <= 'z' && !appeared[m])
        {
            ++m;
        }

        if(n < k)
        {
            char[] t = Arrays.copyOf(s, k);
            for(int i = n; i < k; ++i)
            {
                t[i] = m;
            }
            System.out.println(new String(t));
        }
        else
        {

            char[] t = Arrays.copyOf(s, k);
            for(int i = k - 1; i >= 0; --i)
            {
                char c = t[i];
                ++c;
                while(c <= 'z' && !appeared[c])
                {
                    ++c;
                }
                if(c <= 'z')
                {
                    t[i] = c;
                    for(int j = i + 1; j < k; ++j)
                    {
                        t[j] = m;
                    }
                    break;
                }
            }
            System.out.println(new String(t));
        }
    }
}
