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.

def get_solution(source, desired_len, alphabet):
	if desired_len <= len(source):
		target = source[:desired_len]
		last_non_terminal = get_last_non_terminal(target, alphabet)
		target = target[:last_non_terminal] + next_char(target[last_non_terminal], alphabet)
	else:
		target = source

	return pad_with_starters(target, desired_len, alphabet)

def get_last_non_terminal(target, alphabet):
	for i, char in reversed(list(enumerate(target))):
		if char != alphabet[-1]:
			return i

def next_char(char, alphabet):
	return alphabet[alphabet.index(char) + 1]

def pad_with_starters(target, desired_len, alphabet):
	return target + (desired_len - len(target)) * alphabet[0]


_, desired_len = [int(x) for x in input().split()]
source = input()
alphabet = list(set(source))
alphabet.sort()

print(get_solution(source, desired_len, alphabet))