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>
#pragma warning(disable : 4786)
#pragma warning(disable : 4996)
template <class T>
inline T bigmod(T p, T e, T M) {
  long long ret = 1LL;
  for (; e > 0LL; e >>= 1LL) {
    if (e & 1LL) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string f, l;
  cin >> f >> l;
  string ans1 = "";
  bool bl = false;
  for (int i = 0, j = 0; i < f.length(); i++) {
    if (i == 0) {
      ans1 += f[i];
      continue;
    }
    if (f[i] >= l[j]) {
      ans1 += l[j];
      bl = true;
      break;
    } else {
      ans1 += f[i];
    }
  }
  if (ans1 == f && bl == false) ans1 += l[0];
  cout << ans1;
}
