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 solve():
    n, k = map(int, input().split())
    S = input()
    L = list(set(list(map(ord,S))))
    L.sort()

    I = {}
    for i, l in enumerate(L):
        I[l] = i

    first = L[0]
    last = L[-1]

    res = ""
    if k > n:
        res += S
        res += chr(first)*(k-n)
        print(res)
        return
    else:
        for i in reversed(range(k)):
            if ord(S[i]) < last:
                remained = L[I[ord(S[i])] + 1]
                res += S[:i]
                res += chr(remained)
                res += chr(first) * (k-i-1)
                print(res)
                return


if __name__ == "__main__":
    solve()


