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.
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF940C extends PrintWriter {
	CF940C() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF940C o = new CF940C(); o.main(); o.flush();
	}

	static final int A = 26;
	void main() {
		int n = sc.nextInt();
		int k = sc.nextInt();
		byte[] ss = sc.next().getBytes();
		boolean[] used = new boolean[A];
		int a;
		for (int i = 0; i < n; i++) {
			a = ss[i] - 'a';
			used[a] = true;
		}
		a = 0;
		while (!used[a])
			a++;
		byte cmin = (byte) (a + 'a');
		a = A - 1;
		while (!used[a])
			a--;
		byte cmax = (byte) (a + 'a');
		ss = Arrays.copyOf(ss, k);
		if (k > n)
			for (int i = n; i < k; i++)
				ss[i] = cmin;
		else {
			int i = k - 1;
			while (ss[i] == cmax)
				ss[i--] = cmin;
			a = ss[i] - 'a' + 1;
			while (!used[a])
				a++;
			ss[i] = (byte) (a + 'a');
		}
		println(new String(ss));
	}
}
