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;
char str[100][21];
char a[100], b[100];
int stlen(char z[]) {
  int i;
  for (i = 0; z[i]; i++)
    ;
  return i;
}
int cmp(int x, int y) {
  int i;
  for (i = 0; str[x][i] && str[y][i]; i++) {
    if (str[x][i] < str[y][i]) return -1;
    if (str[x][i] > str[y][i]) return 1;
  }
  if (str[x][i] == 0 && str[y][i]) return -1;
  if (str[x][i] && str[y][i] == 0) return 1;
  return 0;
}
int main() {
  int i, j, la, lb, k, l;
  scanf("%s%s", a, b);
  la = stlen(a);
  lb = stlen(b);
  for (i = 0; i < la; i++) {
    for (j = 0; j < lb; j++) {
      for (k = 0; k <= i; k++) {
        str[i * lb + j][k] = a[k];
      }
      for (l = 0; l <= j; k++, l++) {
        str[i * lb + j][k] = b[l];
      }
      str[i * lb + j][k] = 0;
    }
  }
  int best = 0;
  for (i = 1; i < la * lb; i++) {
    if (cmp(i, best) < 0) best = i;
  }
  printf("%s\n", str[best]);
  return 0;
}
