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 sys
def rem_xtra(li):
    w=[]
    for i in li:
        if(i not in w):
            w.append(i)
    return w
t=int(raw_input().split(" ")[1])
s=raw_input()
li=sorted(s)
li=rem_xtra(li)
ans=""
if(len(s)<t):
    for i in range(t-len(s)):
        s=s+li[0]
    print s
else:
    s=s[:t]
    w=s
    indi=0
    for i in range(len(s)-1,-1,-1):
        for j in li:
            if(j>s[i]):
                w=w[:i]
                ans=w+j+(li[0]*(len(s)-i-1))
                print ans
                sys.exit()
    if(indi==0):
        print li[0]*t
        sys.exit()
