The preferred way to generate user login in Polygon is to concatenate a prefix of the user's first name and a prefix of their last name, in that order. Each prefix must be non-empty, and any of the prefixes can be the full name. Typically there are multiple possible logins for each person.

You are given the first and the last name of a user. Return the alphabetically earliest login they can get (regardless of other potential Polygon users).

As a reminder, a prefix of a string s is its substring which occurs at the beginning of s: "a", "ab", "abc" etc. are prefixes of string "{abcdef}" but "b" and 'bc" are not. A string a is alphabetically earlier than a string b, if a is a prefix of b, or a and b coincide up to some position, and then a has a letter that is alphabetically earlier than the corresponding letter in b: "a" and "ab" are alphabetically earlier than "ac" but "b" and "ba" are alphabetically later than "ac".

Input

The input consists of a single line containing two space-separated strings: the first and the last names. Each character of each string is a lowercase English letter. The length of each string is between 1 and 10, inclusive. 

Output

Output a single string — alphabetically earliest possible login formed from these names. The output should be given in lowercase as well.

Examples

Input

harry potter


Output

hap


Input

tom riddle


Output

tomr
t = list(input())
n = len(t)
for i in range(n):
    if t[i] == " ":
        s = i+1
        break
j = 0
k = 0
h = s-1
fla = 0
li = []
li.append(t[j])
while( fla == 0):
    if t[j+1]<t[s] and fla == 0 and j<h-1 :
        li.append(t[j+1])
        j = j+1
    elif (t[j+1]>=t[s] and j<h-1 and fla == 0 ):
       
        fla = 1
        li.append(t[s])
    elif j == h-1:
        j = j +1
        li.append(t[s])
        fla = 1
    
    
   
        
     
x= len(li)
for v in range(x):
    print(li[v],end="")