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.
import sys
import io, os
input = sys.stdin.readline

def main():

    n, k = map(int, input().split())
    s = str(input().rstrip())

    x = [ord(c)-ord('a') for c in s]
    A = list(set(x))
    A.sort()
    import heapq
    import bisect

    if k <= n:
        same = ['']
        large = []
        #heapq.heapify(large)
        for i in range(k):
            p = x[i]
            nsame = []
            nlarge = []
            #heapq.heapify(nlarge)
            flag = False
            for t in same:
                j = bisect.bisect_left(A, p)
                if j != len(A)-1:
                    nt = t + chr(A[j]+ord('a'))
                    nsame.append(nt)
                    nt = t + chr(A[j+1]+ord('a'))
                    #heapq.heappush(nlarge, nt)
                    nlarge.append(nt)
                    flag = True
                else:
                    nt = t + chr(A[j]+ord('a'))
                    nsame.append(nt)
            if not flag:
                if large:
                    #t = heapq.heappop(large)
                    t = large[0]
                    nt = t + chr(A[0]+ord('a'))
                    #heapq.heappush(nlarge, nt)
                    nlarge.append(nt)
            same = nsame
            large = nlarge
            if large:
                large.sort()
            #print(same)
            #print(large)
        print(large[0])

    else:
        p = chr(A[0]+ord('a'))
        ans = s+p*(k-n)
        print(ans)

if __name__ == '__main__':
    main()
