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.util.Arrays;
import java.util.NavigableSet;
import java.util.TreeSet;

public class PhoneNumbers {
    public static void main(String[] args) throws Exception {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        String[] firstLine = reader.readLine().split(" ");
        int sourceStringLength = Integer.parseInt(firstLine[0]);
        int resultStringLength = Integer.parseInt(firstLine[1]);
        String sourceMultiset = reader.readLine();
        char[] sourceMultisetChars = sourceMultiset.toCharArray();
        NavigableSet<Character> set = new TreeSet<>();
        for (char c : sourceMultisetChars) {
            set.add(c);
        }
        char[] positions = new char[resultStringLength];
        System.arraycopy(sourceMultisetChars, 0, positions, 0, Math.min(resultStringLength, sourceMultisetChars.length));
        for (int i = sourceStringLength; i < resultStringLength; i++) {
            positions[i] = set.first();
        }
        if (compareTo(positions, sourceMultisetChars) > 0) {
            System.out.println(positions);
            return;
        }
        for (int i = positions.length - 1; i >= 0; i--) {
            while (true) {
                Character higher = set.higher(positions[i]);
                if (higher == null) {
                    break;
                }
                positions[i] = higher;
                if (compareTo(positions, sourceMultisetChars) > 0) {
                    System.out.println(positions);
                    return;
                }
            }
            positions[i] = set.first();
        }
    }

    private static int compareTo(char[] v1, char[] v2) {
        int len1 = v1.length;
        int len2 = v2.length;
        int lim = Math.min(len1, len2);
        int k = 0;
        while (k < lim) {
            char c1 = v1[k];
            char c2 = v2[k];
            if (c1 != c2) {
                return c1 - c2;
            }
            k++;
        }
        return len1 - len2;
    }

    private static char[] toCharArray(int[] intArray) {
        char[] result = new char[intArray.length];
        for (int i = 0; i < intArray.length; i++) {
            result[i] = (char) intArray[i];
        }
        return result;
    }
}
