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 to_int(s):
	return ord(s) - ord('a')

def to_char(i):
	return chr(i + ord('a'))

line = [int(x) for x in input().split()]

n = line[0]
k = line[1]

st = input()

present = [False for i in range(26)]
convert = [0 for i in range(k)]

for i in range(len(st)):
	present[to_int(st[i])] = True

mapper = {}
unmapper = []

cont = 0
for i in range(26):
	if present[i]:
		mapper[to_char(i)] = cont
		unmapper.append(to_char(i))
		cont += 1

base = len(mapper)

if n >= k:
	for i in range(min(k, len(st))):
		convert[i] = mapper[st[i]]

	convert[-1] += 1
	for i in range(k - 1):
		j = k - (i + 1)
		convert[j - 1] += convert[j] // base
		convert[j] = convert[j] % base

	ans = ''
	for i in range(len(convert)):
		ans += unmapper[convert[i]]

	print(ans)

else:
	while len(st) < k:
		st += unmapper[0]

	print(st)