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 fractions import gcd
from math import factorial, ceil, sqrt, atan2, log, pi, e, asin,acos, cos, sin
from itertools import *
from fractions import Fraction
import string
import copy
import random
import bisect
from decimal import *
def id_generator(size=20, chars=string.digits):
    return ''.join(random.choice(chars) for _ in range(size))
 
def mp():
    return map(int,str(raw_input()).split())

n,k=mp()
s=str(raw_input())
l=[]
for i in s:
	l+=[i]
l=list(set(l))
#print l
l.sort()
a=l[0]
d={}
for i in range(len(l)-1):
	d[l[i]]=l[i+1]

if n>=k:
	z=list(s[:k])

	for i in range(len(z)-1,-1,-1):
		if z[i]!=l[-1]:
			z[i]=d[z[i]]
			break
	for j in range(i+1,len(z)):
		z[j]=a

	ans=''
	for i in z:
		ans+=i
	print ans
else:
	print s+a*(k-n)