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.*;
public class Solution
{
	public static void main(String[] args)throws IOException
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		String temp1=br.readLine();
		String[] temp2=temp1.split(" ");
		String fn=temp2[0];
		String ln=temp2[1];
		char[] n1=fn.toCharArray();
		char[] n2=ln.toCharArray();
		int l1=n1.length;
		int l2=n2.length;
		String ans="";
		ans+=n1[0];
		int pi1=1;
		int pi2=0;
		while(pi1<l1&&pi2<l2&&pi2<1)
		{
			if(n1[pi1]<n2[pi2])
			{
				ans+=n1[pi1];
				pi1++;
			}
			else if(n1[pi1]>=n2[pi2])
			{
				ans+=n2[pi2];
				pi2++;
			}
		}
		if(pi2<1)
		{
			ans+=n2[pi2];
		}
		System.out.println(ans);
	}
}