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.
def main():
    n, k = [int(v) for v in input().split()]
    s = input()

    d = {}
    for i in range(n):
        d[s[i]] = i

    alpha = sorted(d.keys())

    if k > n:
        print(s + (k - n) * alpha[0])
        return

    break_pos = None
    break_char = None
    for i in range(k-1, -1, -1):
        if s[i] != alpha[len(alpha) - 1]:
            break_pos = i
            break_char = alpha[alpha.index(s[i]) + 1]
            break

    res = []
    for i in range(k):
        if i < break_pos:
            res.append(s[i])
        elif i == break_pos:
            res.append(break_char)
        else:
            res.append(alpha[0])
    print(''.join(res))


if __name__ == '__main__':
    main()
