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.*;
import java.util.*;

public class C {

	static StringBuilder st = new StringBuilder();

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		int n = sc.nextInt(), k = sc.nextInt();
		String or = sc.next();
		char[] s = or.toCharArray();

		TreeSet<Character> set = new TreeSet<>();

		for (char c : s) set.add(c);

		
		if (n < k) 
		{
			char[] ans = new char[k];
			Arrays.fill(ans, set.first().charValue());

			for (int i = 0; i < Math.min(n, k); i++)
				ans[i] = s[i];

			System.out.println(ans);
			return;
		}
		
		char[] curr = new char[set.size()];
		int [] idxOf = new int [26] ;
		
		int idx = 0;
		
		for (char c : set)
			curr[idxOf[c - 'a'] = idx++] = c;
		
		s = Arrays.copyOf(s, k);
		
		
		for(int i = k - 1 ; i >= 0; i--)
		{
			if(idxOf[s[i] - 'a'] + 1  == set.size())
				s[i] = curr[0];
			else
			{
				s[i] = curr[idxOf[s[i] - 'a'] + 1 ];
				break ; 
			}
		}
		
		out.println(s);
		
		out.flush();
		out.close();

	}

	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

		Scanner(InputStream in) {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() throws Exception {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());

			return st.nextToken();
		}

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

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

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

	}

	static void shuffle(int[] a) {
		int n = a.length;
		for (int i = 0; i < n; i++) {
			int r = i + (int) (Math.random() * (n - i));
			int tmp = a[i];
			a[i] = a[r];
			a[r] = tmp;
		}
	}

}