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;
const long long INF = 1e18;
const int mod = 1e9 + 7;
const int N = 1e5 + 10;
inline int add(int x, int y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
inline int mul(int x, int y) {
  x = (1LL * x * y) % mod;
  return x;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  char s1[100] = {'\0'};
  char s2[100] = {'\0'};
  cin >> s1 >> s2;
  char a[100] = {'\0'};
  a[0] = s1[0];
  int k = 1;
  for (int i = 1; s1[i] != '\0'; i++) {
    char c = s1[i];
    if (s2[0] != '\0' && c < s2[0]) {
      a[k] = s1[i];
      k++;
    } else {
      break;
    }
  }
  a[k] = s2[0];
  for (int i = 0; a[i] != '\0'; i++) {
    cout << a[i];
  }
}
