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.
from sys import stdin, stdout
ti = lambda : stdin.readline().strip()
ma = lambda fxn, ti : map(fxn, ti.split())
ol = lambda arr : stdout.write(' '.join(str(i) for i in arr) + '\n')
os = lambda i : stdout.write(str(i) + '\n')
olws = lambda arr : stdout.write(''.join(str(i) for i in arr) + '\n')


n, k = ma(int, ti())
s = ti()

letters = "abcdefghijklmnopqrstuvwxyz"
d = {}
for i in range(26):
	d[letters[i]] = i

arr = [False] * 26
for i in range(n):
	arr[d[s[i]]] = True


first_letter = None
first_letter_lexi = None
for i in range(26):
	if arr[i]:
		first_letter_lexi = i
		first_letter = letters[i]
		break
first_letter = letters[i]

if k > n:
	news = s[:k]
	news += (k-n)*first_letter
else:
	news = s[:k]
	i = k-1
	current_letter_lexi = d[news[i]]
	current_letter = news[i]

	last_letter = letters[25]
	last_letter_lexi = 25
	for i in range(25, -1, -1):
		if arr[i]:
			break
		last_letter_lexi -= 1
	last_letter = letters[last_letter_lexi]

	i = k-1
	while current_letter_lexi >= last_letter_lexi:
		i -= 1
		current_letter_lexi = d[news[i]]

	just_bigger_letter = None
	just_bigger_letter_lexi = None
	for j in range(current_letter_lexi+1, 26):
		if arr[j]:
			just_bigger_letter_lexi = j
			just_bigger_letter = letters[j]
			break

	news = news[:i] + just_bigger_letter + (k-i-1)*first_letter

os(news)