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.lang.*;
import java.math.*;
import java.util.*;

public class C {
	public static void main(String[] args) {
		FastReader fr = new FastReader();
		int n = fr.nextInt();
		int k = fr.nextInt();
		String str = fr.next();
		TreeSet<Character> ts = new TreeSet<>();
		char[] carr = new char[k+1];
		for(int i = 0; i < k; i++){
            if(i < str.length()) carr[i] = str.charAt(i);
            else carr[i] = 'a'-1;
        }
        carr[k] = 'z';
		for(int i = 0; i < n; i++){
			ts.add(str.charAt(i));
		}
        char[] carr2 = new char[k+1];
        int ii = k;
        while(carr[ii] >= ts.last()){
            carr2[ii] = ts.first();
            ii--;
        }
        carr2[ii] = ts.higher(carr[ii]);
        ii--;
        for(; ii >= 0; ii--){
            carr2[ii] = carr[ii] < ts.first() ? ts.first() : carr[ii];
        }
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i < k; i++){
            sb.append(carr2[i]);
        }
        System.out.println(sb);
	}
}

class FastReader{
	BufferedReader br;
	StringTokenizer st;

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

	String next(){
		while(st==null || !st.hasMoreTokens()){
			try{
				st = new StringTokenizer(br.readLine());
			}catch(IOException e){
				e.printStackTrace();
			}
		}
		return st.nextToken();
	}

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

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

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

	String nextLine(){
		String str = "";
		try{
			str = br.readLine();
		}catch(IOException e){
			e.printStackTrace();
		}
		return str;
	}
}

//source: http://www.geeksforgeeks.org/fast-io-in-java-in-competitive-programming/