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.BufferedReader;
import java.io.Closeable;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
import java.util.StringTokenizer;

public class Main {

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
	
		final int n = sc.nextInt();
		final int k = sc.nextInt();
		
		char[] s = sc.next().toCharArray();
		
		boolean[] set = new boolean[26];
		for(char c : s){ set[c - 'a'] = true; }
		
		char smallest = s[0];
		for(char c : s){ smallest = (char) Math.min(smallest, c); }

		char[] t = new char[k];
		
		if(n < k){
			for(int i = 0; i < n; i++){ t[i] = s[i]; }
			for(int i = n; i < k; i++){ t[i] = smallest; }
		}else{
			for(int i = 0; i < k; i++){ t[i] = s[i]; }
			
			LOOP:
			for(int i = k - 1; i >= 0; i--){
				for(int c = (s[i] - 'a' + 1); c < set.length; c++){
					if(!set[c]){ continue; }
					
					t[i] = (char)(c + 'a');
					for(int j = i + 1; j < k; j++){ t[j] = smallest; }
					break LOOP;
				}
			}
		}
		
		System.out.println(String.valueOf(t));
	}

	public static class Scanner implements Closeable {
		private BufferedReader br;
		private StringTokenizer tok;

		public Scanner(InputStream is) throws IOException {
			br = new BufferedReader(new InputStreamReader(is));
		}

		private void getLine() throws IOException {
			while (!hasNext()) {
				tok = new StringTokenizer(br.readLine());
			}
		}

		private boolean hasNext() {
			return tok != null && tok.hasMoreTokens();
		}

		public String next() throws IOException {
			getLine();
			return tok.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public int[] nextIntArray(int n) throws IOException {
			final int[] ret = new int[n];
			for (int i = 0; i < n; i++) {
				ret[i] = this.nextInt();
			}
			return ret;
		}

		public long[] nextLongArray(int n) throws IOException {
			final long[] ret = new long[n];
			for (int i = 0; i < n; i++) {
				ret[i] = this.nextLong();
			}
			return ret;
		}

		public void close() throws IOException {
			br.close();
		}
	}
}