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>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2")
using namespace std;
using vi = vector<long long>;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
long long powmod(long long a, long long b, long long mod) {
  if (b == 0 || a == 1) {
    if (mod == 1)
      return 0;
    else
      return 1;
  }
  if (b % 2 == 0) {
    long long k = powmod(a, b / 2, mod);
    return (k * k) % mod;
  } else {
    long long k = powmod(a, b / 2, mod);
    return ((k * k) % mod * a) % mod;
  }
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (a > b)
    return gcd(a % b, b);
  else
    return gcd(b % a, a);
}
long long prime(long long p) {
  if (p == 1) return 0;
  for (long long i = 2; i * i <= p; i++) {
    if (p % i == 0 && i < p) return i;
  }
  return 1;
}
long long inv(long long a, long long mod) { return powmod(a, mod - 2, mod); }
long long random_modul() {
  uniform_int_distribution<long long> u1(1e9, 2e9);
  long long pepega = u1(rnd);
  while (prime(pepega) != 1) pepega = u1(rnd);
  return pepega;
}
template <typename T1>
void ifmax(T1& max, T1 kek) {
  if (kek > max) max = kek;
  return;
}
template <typename T1>
void ifmin(T1& min, T1 kek) {
  if (kek < min) min = kek;
  return;
}
template <typename T1>
istream& operator>>(std::istream& in, pair<T1, T1>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1>
istream& operator>>(std::istream& in, vector<T1>& a) {
  for (long long i = 0; i < a.size(); i++) in >> a[i];
  return in;
}
template <typename T1>
ostream& operator<<(std::ostream& out, vector<T1>& b) {
  for (long long i = 0; i < b.size(); i++) out << b[i];
  return out;
}
template <typename T1>
void sort(vector<T1>& m) {
  sort(m.begin(), m.end());
}
template <typename T1>
void reverse(vector<T1>& m) {
  reverse(m.begin(), m.end());
}
long long mod = 1000000007;
long long fac[200005];
long long inv_fac[200005];
long long co(long long a, long long b) {
  long long ans = 1;
  ans *= fac[b];
  ans *= inv_fac[b - a];
  ans %= mod;
  ans *= inv_fac[a];
  return ans % mod;
}
void solve() {
  string a, b;
  cin >> a >> b;
  long long s1 = a.size();
  long long s2 = b.size();
  string ans;
  for (long long i = 1; i <= s1; i++) {
    for (long long j = 1; j <= s2; j++) {
      string cur = "";
      for (long long k = 0; k < i; k++) cur += a[k];
      for (long long k = 0; k < j; k++) cur += b[k];
      if (i == 1 && i == j)
        ans = cur;
      else
        ans = min(ans, cur);
    }
  }
  cout << ans;
  return;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  long long multitest = 0;
  long long tututu;
  tututu = 1;
  long long combinatorics = 0;
  if (combinatorics) {
    fac[1] = 1;
    fac[0] = 1;
    for (long long i = 2; i < 200005; i++) {
      fac[i] = fac[i - 1] * i;
      fac[i] %= mod;
    }
    for (long long i = 0; i < 200005; i++) {
      inv_fac[i] = inv(fac[i], mod);
    }
  }
  if (multitest) cin >> tututu;
  for (long long qwerty = 0; qwerty < tututu; qwerty++) solve();
  return 0;
}
