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 CFR466C {
    static int n, k;
    static String s, ins[];
    static boolean a[] = new boolean[26], isFind = false;
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        ins = br.readLine().split(" ");
        n = Integer.parseInt(ins[0]);
        k = Integer.parseInt(ins[1]);
        s = br.readLine();
        for (int i = 0; i < s.length(); i++) {
            a[s.charAt(i) - 97] = true;
        }
        StringBuilder sb = new StringBuilder();

        if (n < k) {
            sb.append(s);
            int t = -1;
            for (int i = 0; i < 26; i++) {
                if (a[i]) {
                    t = i; break;
                }
            }
            for (int i = 0; i < k - n; i++) {
                sb.append((char)(t + 97));
            }
            sb.append("\n");
        }else {
            char[] ca = s.toCharArray();
            int idx = k - 1;
            while (true) {
                boolean isBigger = false;
                int t = -1;
                for (int i = ca[idx] - 96; i < 26; i++) {
                    if (a[i]) {
                        t = i;
                        isBigger = true; break;
                    }
                }
                if (isBigger) {
                    ca[idx] = (char)(t + 97);
                    break;
                }else {
                    int first = -1;
                    for (int i = 0; i < 26; i++) {
                        if (a[i]) {
                            first = i; break;
                        }
                    }
                    ca[idx--] = (char)(first + 97);
                }
            }
            for (int i = 0; i < k; i++) {
                sb.append(ca[i]);
            }
            sb.append("\n");
        }



        bw.write(sb.toString());
        bw.flush();bw.close();
    }

}
