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.util.Scanner;

public class A {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		
		String fname = in.next();
		String sname = in.next();
		
		String ans = "";
		
		for(int i = 0; i < fname.length(); ++i) {
			String pref = fname.substring(0, i+1);
			for(int j = 0; j < sname.length(); ++j) {
				String sfix = sname.substring(0, j+1);
				
				String test = pref + sfix;
				
				if(comp(ans,test)) {
					ans = test;
				}
			}
		}
		
		System.out.println(ans);
	}
	
	static boolean comp(String a, String t) {
		if(a == "") return true;
		else {
			int i = 0;
			int j = 0;
			
			while(i < a.length() && j < t.length()) {
				if(a.charAt(i) < t.charAt(j)) return false;
				if(a.charAt(i) > t.charAt(j)) return true;
				i++;
				j++;
			}
			if(i == a.length() && j < t.length()) return false;
			else return true;
		}
	}

}
