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.Scanner;
public class Main {
    public static void main(String args[]) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        int k = s.nextInt();
        String str = s.next();
        int min = 10000;
        int max = -1;
        boolean present[] = new boolean[26];
        for(int i = 0; i < n; i++){
            min = Math.min(min,str.charAt(i) - 'a');
            max = Math.max(max,str.charAt(i) - 'a');   
            present[str.charAt(i) - 'a'] = true;
        }
        if(k > n){
            System.out.print(str);
            int diff = k - n;
            while(diff > 0){
                System.out.print((char)('a' + min));
                diff--;
            }
            return;
        }
        char c = (char)(min);
        char arr[] = new char[k];
        int start = k - 1;
        while(start >= n){
            arr[start] = c;
            start--;
        }
        while(start >= 0 && str.charAt(start) - 'a' == max){
            start--;
        }
        for(int i = str.charAt(start) - 'a' + 1; i < 26; i++){
            if(present[i] == true){
                arr[start] = (char)('a' + i);
                break;
            }
            
        }
        int right = start + 1,left = start - 1;
        while(right < Math.min(n,k)){
            arr[right] = (char)(min + 'a');
            right++;

        }
        while(left >= 0){
            arr[left] = str.charAt(left);
            left--;
        }
        for(int i = 0; i < k; i++){
            System.out.print(arr[i]);
        }
    }
}