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 PhoneNumbers {
  public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    System.out.println(solution(in.nextInt(), in.nextInt(), in.next()));
  }
  
  public static String solution(int n, int k, String s) {
    char[] sChars = s.toCharArray();
    Set<Character> charSet = getCharSet(sChars);
    char[] res = new char[k];
    
    for (int i = 0; i < Math.min(n, k); i++) {
      res[i] = sChars[i];
    }
    
    if (n < k) {
      updateRemainderWithSmallest(res, n, charSet);
      return new String(res);
    }
    
    for (int i = k - 1; i >= 0; i--) {
      char nextLargest = getNextLargest(res[i], charSet);
      if (nextLargest != '0') {
        res[i] = nextLargest;
        updateRemainderWithSmallest(res, i + 1, charSet);
        return new String(res);
      }
    }
    
    return "";
  }
  
  private static Set<Character> getCharSet(char[] s) {
    Set<Character> toReturn = new HashSet<Character>();
    for (char c : s) {
      toReturn.add(c);
    }
    return toReturn;
  }
  
  private static char getNextLargest(char c, Set<Character> charSet) {
    for (char i = (char) (c + 1); i <= 'z'; i++) {
      if (charSet.contains(i)) {
        return i;
      }
    }
    return '0';
  }
  
  private static void updateRemainderWithSmallest(char[] res, int i, Set<Character> charSet) {
    char smallest = 'z';
    for (char c = 'a'; c < 'z'; c++) {
      if (charSet.contains(c)) {
        smallest = c;
        break;
      }
    }
    
    for (int j = i; j < res.length; j++) {
      res[j] = smallest;
    }
  }
}