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 sys import stdin as cin

def main():
    n, m = map(int, input().split())
    s = input()
    q = set()
    c = 'z'
    x = 'a'
    for i in s:
        q.add(i)
        if ord(c) > ord(i):
            c = i
        if ord(x) < ord(i):
            x = i
    if n >= m:
        s = s[:m]
        for i in range(m - 1, -1, -1):
            if s[i] != x:
                print(s[:i], end = '')
                w = 1+ord(s[i])
                while chr(w) not in q:
                    w += 1
                print(chr(w) + c * (m - i - 1))
                return
##        for i in range(m):
##            if i < n and s[i] in q:
##                q.remove(s[i])
##                if len(q) == 1:
##                    print(*q, end = '')
##                elif len(q) == 0:
##                    print(c, end = '')
##                else:
##                    print(s[i], end = '')
##            else:
##                if i < n and len(q) <= 1:
##                    print(c, end = '')
##                else:
##                    print(s[i], end = '')
    else:
        print(s + c * (m - n))

main()
