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.*;

public class c466 {

    public static char findMin(int[] a) {
	for(int i = 0;i <  26;++i)
	    if(a[i] > 0)
		return (char)('a'+i);
	return 'A';
    }

    public static char findNext(int[] a,char ch) {
	for(int i = ch-'a'+1;i < 26;++i)
	    if(a[i] > 0)
		return (char)('a'+i);
	return 'A';
    }
    
    public static void main(String[] args) {
	Scanner sc = new Scanner(System.in);
	int n = sc.nextInt(),k = sc.nextInt();
	String s = sc.next();

	int[] a = new int[26];

	for(int i = 0;i < s.length();++i)
	    a[s.charAt(i)-'a'] ++;

	char mi = findMin(a);

	if(k > n) {
	    System.out.print(s);

	    for(int i = n+1;i <= k;++i)
		System.out.print(mi);
	    System.out.println();
	}

	else {
	    for(int i = k-1;i >= 0;--i) {
		if(findNext(a,s.charAt(i)) != 'A'){
		    System.out.print(s.substring(0,i)+findNext(a,s.charAt(i)));
		    for(int j = i+1;j < k;++j)
			System.out.print(mi);
		    System.out.println();
		    break;
		}
	    }
	}
	
    }
}
