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.
s=input()
d=list(map(int,s.split(' ')))
l=d[0]
c=m=d[1]
t=input()
s=[]
p=[]
o=[]
for h in range(0,l):
    p.append(t[h])
    if(t[h] not in o):
        o.append(t[h])
o.sort()
def polo(r):
    for y in range(0,len(o)):
        if(t[r]==o[y]):
            return(o[y+1])
            break
if(l<m):
    t=t+o[0]*(m-l)
    print(t)
else:
    s=p[0:m-1]
    if(t[m-1]!=o[len(o)-1]):
        s.append(polo(m-1))
        s=''.join(map(str,s))
    else:
        s.append('k')
        while(t[m-1]==o[len(o)-1]):
            del(s[m-1])
            m=len(s)
        del(s[m-1])
        s.append(polo(m-1))
        s=''.join(map(str,s))
        s=s+o[0]*(c-m)
    print(s)
