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.*;
import java.io.*;

public class B {

	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		String[] ree = in.readLine().split("\\s++");
		int n = Integer.parseInt(ree[0]);
		int k = Integer.parseInt(ree[1]);
		int excess = k - n;
		boolean cheat = false;
		if(k > n) cheat = true;
		String inp = in.readLine();
		char[] arrS = inp.toCharArray();
		char min = 'z';
		char max = 'a';
		for(int i = 0; i < arrS.length; i++) {
			if(arrS[i] < min) min = arrS[i];
			if(arrS[i] > max) max = arrS[i];
		}
		if(cheat) {
			System.out.print(inp);
			for(int i = 0; i < excess; i++) System.out.print(min);
			System.out.println();
			return;
		}
		char[] arr = inp.substring(0, k).toCharArray();
		for(int i = arr.length - 1; i >= 0; i--) {
			if(arr[i] == max) {
				arr[i] = min;
			} else {
				char replace = 'z';
				for(int j = 0; j < arrS.length; j++) {
					/*System.out.println(replace + " " + arrS[j] + " to replace " + arr[i]);
					System.out.println(arr[j] < replace);
					System.out.println(arrS[j] > arr[i]);*/
					if(arrS[j] < replace && arrS[j] > arr[i]) {
						//System.out.println("yas");
						replace = arrS[j];
					}
				}
				arr[i] = replace;
				break;
			}
		}
		System.out.println(new String(arr));
	}
}