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
''' ===============================
    -- @uthor : Kaleab Asfaw
    -- Handle : kaleabasfaw2010
    -- Bio    : High-School Student
    ==============================='''

# Fast IO
import sys
import os
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
    def __init__(self, file): self._fd = file.fileno(); self.buffer = BytesIO(); self.writable = "x" in file.mode or "r" not in file.mode; self.write = self.buffer.write if self.writable else None
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b: break
            ptr = self.buffer.tell(); self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0; return self.buffer.read()
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)); self.newlines = b.count(b"\n") + (not b); ptr = self.buffer.tell(); self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1; return self.buffer.readline()
    def flush(self):
        if self.writable: os.write(self._fd, self.buffer.getvalue()); self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
    def __init__(self, file): self.buffer = FastIO(file); self.flush = self.buffer.flush; self.writable = self.buffer.writable; self.write = lambda s: self.buffer.write(s.encode("ascii")); self.read = lambda: self.buffer.read().decode("ascii"); self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout); input = lambda: sys.stdin.readline().rstrip("\r\n")
# Others
# from math import floor, ceil, gcd
# from decimal import Decimal as d
mod = 10**9+7
def lcm(x, y): return (x * y) / (gcd(x, y))
def fact(x, mod=mod):
	ans = 1
	for i in range(1, x+1): ans = (ans * i) % mod
	return ans
def arr2D(n, m, default=0):
	lst = []
	for i in range(n): temp = [default] * m; lst.append(temp)
	return lst
def sortDictV(x): return {k: v for k, v in sorted(x.items(), key = lambda item : item[1])}

def solve(a, b):
	i, j = 0, 0
	ans = ""
	while i < len(a) or j < len(a):
		if i == 0 or (i < len(a) and ord(a[i]) < ord(b[j])):
			ans += a[i]
			i += 1
		else:
			ans += b[j]
			break
	return ans


a, b = input().split()
print(solve(a, b))