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.
n, k = [int(x) for x in input().split()]
S = input()
cset = set(S)
clist = list(cset)
clist.sort()

# print(cset)
# print(clist)

def lexnext(Si):
    for c in clist:
        if c > Si:
            return c

if n < k:
    print(S + clist[0]*(k-n))
else:
    T = list(S[:k])
    for i in range(k-1, -1, -1):
        if lexnext(T[i]) is not None:
            T[i] = lexnext(T[i])
            for j in range(i+1, k):
                T[j] = clist[0]
            break
    print("".join(T))


# # print(lexnext("a"))
# # print(lexnext("b"))
# # print(lexnext("c"))

# res = []
# flag = False
# for i in range(1, min(k,n)):
#     if S[i-1] in cset:
#         res.append(S[i-1])
#     else:
#         res.append(lexnext(S[i-1]))
#         for j in range(i, min(k,n)):
#             res.append(clist[0])
#         flag = True
#         break

# if flag:
#     res.append(clist[0])
# else:
#     res.append(lexnext(S[k-1]))

# if k >= n:
#     for i in range(n+1, k+1):
#         res.append(clist[0])

# print(res)

# res = []
# for i in range(1, k):
#     if i == k:
#         for c in clist:
#             if c > S[i]
#     if S[i] in cset:
#         res.append(S[i])

