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;
bool matsolve(int i, int j, int n, int m) {
  if (i - 1 >= 0) {
    if (j - 1 >= 0) {
    }
    if (j + 1 < m) {
    }
  }
  if (i + 1 < n) {
    if (j - 1 >= 0) {
    }
    if (j + 1 < m) {
    }
  }
  if (j - 1 >= 0) {
  }
  if (j + 1 < m) {
  }
}
int main() {
  string s, s1;
  cin >> s >> s1;
  int a_ptr = 1, b_ptr = 0;
  string ans = "";
  ans.push_back(s[0]);
  while (a_ptr < (int)(s.size()) && s[a_ptr] < s1[b_ptr]) {
    ans.push_back(s[a_ptr]);
    a_ptr++;
  }
  char last_b = '1';
  while (b_ptr < (int)(s1.size()) &&
         (a_ptr == (int)(s.size()) || s[a_ptr] >= s1[b_ptr])) {
    last_b = s1[b_ptr];
    ans.push_back(s1[b_ptr]);
    if (a_ptr <= (int)(s.size()) || s[a_ptr] > s1[b_ptr]) break;
    b_ptr++;
  }
  cout << ans;
}
