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.Scanner;

public class PhoneNumbers {
    public static void main(String args []) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        sc.nextLine();
        char s[] = sc.nextLine().toCharArray();
        System.out.println(new PhoneNumbers(s).solve(k));
    }
    
    private final int N = 'z' - 'a' + 1;
    private final int next[] = new int[N];
    private int smallest = -1;
    private final char s[];
    
    PhoneNumbers(char s[]) {
        this.s = s;
        boolean used[] = new boolean[N];
        for (char c : s) used[c - 'a'] = true;
        int prev = -1;
        for (int i = 0; i < N; i++) {
            next[i] = -1;
            if (used[i]) {
                if (prev != -1) {
                    next[prev] = i;
                } else {
                    smallest = i;
                }
                prev = i;
            }
        }
    }
    
    char[] solve(int k) {
        char res[] = java.util.Arrays.copyOf(s, k);
        if (k > s.length) {
            expand(res, k);
        } else {
            increment(res, k);
        }
        return res;
    }
    
    void expand(char res[], int k) {
        for (int i = s.length; i < k; i++) res[i] = (char) ('a' + smallest);
    }
    
    void increment(char res[], int k) {
        for (int i = k-1; i >=0; i--) {
            int nx = next[res[i] - 'a'];
            if (nx == -1) {
                res[i] = (char) ('a' + smallest);
            } else {
                res[i] = (char) ('a' + nx);
                break;
            }
        }
    }
}