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
# May the Speedforce be with us...
'''
for _ in range(int(input())):
arr=list(map(int,input().split()))
n,k=map(int,input().split())
n=int(input())
s=input()

from collections import defaultdict
d=defaultdict()

d=dict()
s=set()
s.intersection()
s.union()
s.difference()


problem statement achhe se padhna hai
age se bhi dekhna hai, piche se bhi dekhna hai
'''
from math import gcd,ceil
from collections import defaultdict as dd
def lcm(a,b):
	return (a*b)//gcd(a,b)
def inin():
	return int(input())
def inar():
	return list(map(int,input().split()))
def ar(element,size):
	return [element for i in range(size)]
def digitsum(num):
	su=0
	while(num):
		su+=num%10
		num//=10
	return su

n,t=input().split()
title=t[0]
ans=n[0]
for i in n[1:]:
	if i>=title:
		break
	ans+=i
ans+=title
print(ans)