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 greater_than(a,cnt):
    for i in range(ord(a)-97+1,26):
        if cnt[i]!=0:
            return chr(i+97)
def smallest(cnt):
    for i in range(26):
        if cnt[i]!=0:
            return chr(i+97)
def last(cnt):
    for i in range(25,-1,-1):
        if cnt[i]!=0:
            return chr(i+97)
        
        
        
n,k = map(int,input().split()) 

s=input()

cnt=[0]*26
for i in range(n):
    cnt[ord(s[i])-97]+=1 
ind1=-1 
ind2=-1   
# 
#if k==n:
#    ans=list(s)
#    for i in range(n-1,-1,-1):
#        if s[i]!=s[i-1]:
#            ind1=i
#            ind2=i-1 
#            break
#    if s[i-1]<s[i]:
#        a = greater_than(s[i-1],cnt)
#        ans[ind2]=a 
#        p=smallest(cnt)
#        for j in range(ind1,n):
#            ans[j]=p
#    else:
#        a=greater_than(str[i],cnt)
#        ans[ind1]=a 
#        for j in range(ind1,n):
#            ans[j]=p
#        
#    print("".join(ans))
if k<=n:
    ans=list(s[:k])
    l=last(cnt)
    c=""
    ind = -1
    for j in range(k-1,-1,-1):
        
        if s[j]!=l:
            c=s[j]
            ind=j
            break
    ans[ind]=greater_than(c,cnt)
    s1=smallest(cnt)
    for j in range(ind+1,k):
        ans[j]=s1
    print("".join(ans))
elif (k>n):
    ans=list(s)
    s1=smallest(cnt)
    for j in range(n,k):
        ans.append(s1)
    print("".join(ans))
        
            
            
            
        
    


            
        
        
   
        
    


    