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;
int n, m, k;
string s[1009];
int ans[1009][1009], vis[1009][1009];
void fun(int x, int y) {
  if (x >= 0 && x < n && y >= 0 && y < m) {
    for (int i = x + 1; i < min(n, x + k + 1); i++) {
      if (s[i][y] == '#' || vis[i][y] == 1 ||
          (ans[i][y] >= 0 && ans[i][y] <= ans[x][y]))
        break;
      ans[i][y] = (ans[x][y] + 1);
      vis[i][y] = 1;
      fun(i, y);
    }
    for (int i = y + 1; i < min(m, y + k + 1); i++) {
      if (s[x][i] == '#' || vis[x][i] == 1 ||
          (ans[x][i] >= 0 && ans[x][i] <= ans[x][y]))
        break;
      ans[x][i] = (ans[x][y] + 1);
      vis[x][i] = 1;
      fun(x, i);
    }
    for (int i = x - 1; i >= max(0, x - k); i--) {
      if (s[i][y] == '#' || vis[i][y] == 1 ||
          (ans[i][y] >= 0 && ans[i][y] <= ans[x][y]))
        break;
      ans[i][y] = (ans[x][y] + 1);
      vis[i][y] = 1;
      fun(i, y);
    }
    for (int i = y - 1; i >= max(0, y - k); i--) {
      if (s[x][i] == '#' || vis[x][i] == 1 ||
          (ans[x][i] >= 0 && ans[x][i] <= ans[x][y]))
        break;
      ans[x][i] = (ans[x][y] + 1);
      vis[x][i] = 1;
      fun(x, i);
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    string s, s1;
    cin >> s >> s1;
    int i;
    for (i = 1; i < s.size(); i++) {
      if (s[i] >= s1[0]) break;
    }
    if (i == s.size())
      cout << s << s1[0];
    else
      cout << s.substr(0, i) << s1[0];
  }
}
