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 collections import defaultdict


def calc(s, n, k):
    input_set = set(s)
    next_el = defaultdict(str)
    input_list = sorted(list(input_set))
    for ndx, value in enumerate(input_list):
        if ndx < len(input_list) - 1:
            next_el[value] = input_list[ndx + 1]
    if k <= len(s):
        t = s[:k]
    else:
        ans = s + min(input_set) * (k - n)
        print ans
        return
    t = list(t)
    i = k - 1
    while i >= 0:
        chr = t[i]
        if chr in next_el:
            t[i] = next_el[chr]
            break
        else:
            t[i] = min(input_set)
        i = i - 1

    print "".join(t)


n, k = map(int, raw_input().strip().split())
s = raw_input().strip()
calc(s, n, k)