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
import java.io.*;
import java.util.*;
public class codeforces {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder str = new StringBuilder();
        PrintWriter out = new PrintWriter(System.out);
        // int n=Integer.parseInt(br.readLine());
        String[] l=br.readLine().split(" ");
        String fname=l[0],lname=l[1];
        String s=String.valueOf(fname.charAt(0));
        l:
        for(int i=1;i<fname.length();i++){
            if(fname.charAt(i)<lname.charAt(0))
            s+=fname.charAt(i);
            else
            {
                // s+=lname.charAt(0);
                break l;
            }
        }
        // if(s.length()==fname.length() && s.charAt(s.length()-1)!=lname.charAt(0))
        s+=lname.charAt(0);
        str.append(s);
        out.print(str+"\n");
        out.close();
    }
}