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;
inline int read() {
  int fh = 1, x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return fh * x;
}
inline void write(const int x) {
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
inline int _abs(const int x) { return x >= 0 ? x : -x; }
inline int _max(const int x, const int y) { return x >= y ? x : y; }
inline int _min(const int x, const int y) { return x <= y ? x : y; }
inline int _gcd(const int x, const int y) { return y ? _gcd(y, x % y) : x; }
inline int _lcm(const int x, const int y) { return x * y / _gcd(x, y); }
char a[100005], b[100005], ans[100005];
int lena, lenb;
signed main() {
  scanf("%s %s", a, b);
  ans[0] = a[0];
  int i;
  for (i = 1; i < strlen(a); i++)
    if (a[i] < b[0])
      ans[i] = a[i];
    else
      break;
  ans[i] = b[0];
  printf("%s\n", ans);
  return 0;
}
