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 time
import math

debug = False

#nWord1Len = int(input())
nWord1Len, nWord2Len = map(int, input().split())
strWord1 = input()
#nWord2Len = int(input())

if debug : print("---")
begin = time.time()

Chars = set()
for char in strWord1: Chars.add(char)

Alphabet = list(Chars)
Alphabet.sort()

strWord2 = ""

if nWord2Len > nWord1Len:
	strWord2 = strWord1 + Alphabet[0] * (nWord2Len - nWord1Len)
else:
	strWord2 = strWord1[:nWord2Len]
	for i in range(nWord2Len,0,-1):
		CurrChar = strWord2[i-1]
		CharIndex = Alphabet.index(CurrChar)
		if CharIndex < len(Alphabet) - 1:
			strWord2 = strWord2[:i-1] + Alphabet[CharIndex+1] + Alphabet[0]*(nWord2Len-i)
			break

if debug : print(time.time() - begin)
if debug : print("---")

print(strWord2)

