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.
#complexity - O(n)
import string

n, k = map(int, input().split())
s = input()
set_of_letters = set(s)

def find_smallest():
  for i in string.ascii_lowercase:
    if i in set_of_letters:
      return i
  return None #should not occur given the conditions

def find_next_greater(letter):
  counter = 1
  while counter < 26:
    next_char = chr((ord(letter)-ord('a') + counter)%26+ord('a'))
    if next_char in set_of_letters:
      return next_char
    counter += 1
  return None
  
if k > n:
  print(s + find_smallest()*(k-n))
else:
  s_list = list(s[:k])
  while k > 0:
    next_greater = find_next_greater(s[k-1])
    s_list[k-1] = next_greater
    if next_greater > s[k-1]: #really greater
      break #should break before k = 0 given the conditions
    k -= 1
  print(''.join(s_list))
