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[] str = sc.nextLine().toCharArray();
		sc.close();
		char[] ans = new char[k];
		int max = str[0];
		int min = str[0];
		for (int i = 0; i < str.length; i++) {
			max = Math.max(str[i], max);
			min = Math.min(str[i], min);

		}
		if (k <= n) {
			int l = str.length;
			for (int i = k - 1; i >= 0; i--) {
				if (str[i] != max) {
					l = i;
					break;
				}
			}
			int minOverl = max;
			for (int i = 0; i < str.length; i++) {
				if (str[i] > str[l]) {
					minOverl = Math.min(str[i], minOverl);
				}
			}
			for (int i = 0; i < l; i++) {
				ans[i] = str[i];
			}
			ans[l] = (char) minOverl;
			for (int i = l + 1; i < k; i++) {
				ans[i] = (char) min;
			}
		} else {// k>n
			for (int i = 0; i < n; i++) {
				ans[i] = str[i];
			}
			for (int i = n; i < k; i++) {
				ans[i] = (char) min;
			}
		}
		StringBuilder out = new StringBuilder();
		for (int i = 0; i < ans.length; i++) {
			out.append(ans[i]);
		}
		System.out.println(out);
	}
}
