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
#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
  os << '{';
  for (int i = (0), __i = (((int)(vec).size())); i < __i; i++) {
    os << vec[i];
    if (i + 1 != ((int)(vec).size())) os << ',';
  }
  os << '}';
  return os;
}
template <class T1, class T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &par) {
  os << '(' << par.first << ',' << par.second << ')';
  return os;
}
int gcd(int x, int y) { return y ? gcd(y, x % y) : abs(x); }
template <class T>
T sqr(T x) {
  return x * x;
}
const long long MOD = 1e9 + 7;
long long bin_pow(long long base, long long p) {
  if (p == 1) return base;
  if (p % 2 == 0) {
    long long t = bin_pow(base, p / 2);
    return t * t % MOD;
  } else
    return bin_pow(base, p - 1) * base % MOD;
}
long long inverse_element(long long x) { return bin_pow(x, MOD - 2); }
long long add(long long a, long long b) { return (a + b) % MOD; }
long long sub(long long a, long long b) { return (MOD + a - b) % MOD; }
long long divide(long long a, long long b) {
  return a * inverse_element(b) % MOD;
}
long long mult(long long a, long long b) { return (a * b) % MOD; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  string s;
  getline(cin, s);
  int surname = 0;
  for (int i = (0), __i = (s.length()); i < __i; i++)
    if (s[i] == ' ') {
      surname = i + 1;
      break;
    }
  vector<string> vec;
  for (int i = (1), __i = (surname); i < __i; i++) {
    for (int j = (1), __j = (s.length() - surname + 1); j < __j; j++) {
      vec.push_back(s.substr(0, i) + s.substr(surname, j));
    }
  }
  sort((vec).begin(), (vec).end());
  cout << vec[0];
  return 0;
}
