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 A[101], B[101];
char ans[101];
char tmp[101];
int judge(int x, int y) {
  int nw = 1;
  while (nw <= x && nw <= y) {
    if (tmp[nw] != ans[nw]) return tmp[nw] < ans[nw];
    nw++;
  }
  if (x <= y) return 1;
  return 0;
}
int main() {
  scanf("%s", A + 1);
  scanf("%s", B + 1);
  int n = strlen(A + 1), m = strlen(B + 1);
  int vis = 0;
  int lena;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (!vis) {
        int tp = 0;
        for (int k = 1; k <= i; k++) ans[++tp] = A[k];
        for (int k = 1; k <= j; k++) ans[++tp] = B[k];
        vis = 1;
        lena = tp;
      } else {
        int tp = 0;
        for (int k = 1; k <= i; k++) tmp[++tp] = A[k];
        for (int k = 1; k <= j; k++) tmp[++tp] = B[k];
        if (judge(tp, lena)) {
          lena = tp;
          for (int k = 1; k <= tp; k++) ans[k] = tmp[k];
        }
      }
    }
  }
  for (int i = 1; i <= lena; i++) printf("%c", ans[i]);
  return 0;
}
