Problem:
Score : 200 points

Problem Statement

You are given strings s and t, consisting of lowercase English letters.
You will create a string s' by freely rearranging the characters in s.
You will also create a string t' by freely rearranging the characters in t.
Determine whether it is possible to satisfy s' < t' for the lexicographic order.

Notes

For a string a = a_1 a_2 ... a_N of length N and a string b = b_1 b_2 ... b_M of length M, we say a < b for the lexicographic order if either one of the following two conditions holds true:

N < M and a_1 = b_1, a_2 = b_2, ..., a_N = b_N.

There exists i (1 \leq i \leq N, M) such that a_1 = b_1, a_2 = b_2, ..., a_{i - 1} = b_{i - 1} and a_i < b_i. Here, letters are compared using alphabetical order.

For example, xy < xya and atcoder < atlas.

Constraints

The lengths of s and t are between 1 and 100 (inclusive).

s and t consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s
t

Output

If it is possible to satisfy s' < t', print Yes; if it is not, print No.

Sample Input 1

yx
axy

Sample Output 1

Yes

We can, for example, rearrange yx into xy and axy into yxa. Then, xy < yxa.

Sample Input 2

ratcode
atlas

Sample Output 2

Yes

We can, for example, rearrange ratcode into acdeort and atlas into tslaa. Then, acdeort < tslaa.

Sample Input 3

cd
abc

Sample Output 3

No

No matter how we rearrange cd and abc, we cannot achieve our objective.

Sample Input 4

w
ww

Sample Output 4

Yes

Sample Input 5

zzz
zzz

Sample Output 5

No

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner scan = new Scanner(System.in);
		String s = scan.next();
		String t = scan.next();
		Character[] sc = new Character[s.length()];
		Character[] tc = new Character[t.length()];
		int length=Math.min(s.length(), t.length());
		for(int i=0;i<s.length();i++) {
			sc[i]=s.charAt(i);
		}
		for(int i=0;i<t.length();i++) {
			tc[i]=t.charAt(i);
		}
		Arrays.sort(sc);
		Arrays.sort(tc, Collections.reverseOrder());
		//System.out.println(Arrays.asList(sc));
		//System.out.println(Arrays.asList(tc));

/*
		boolean hantei = true;
		int count = 0;
		for(int i=0;i<length;i++) {
			int sci=sc[i];
		//	System.out.println(sci);
			int tci=tc[i];
		//	System.out.println(tci);
			if(sci<tci) {
				hantei=true;
				break;
			}else if(tci<sci) {
				hantei =false;
				break;
			}else {
				count++;
			}
		}
//System.out.println(count);
		//if(hantei==true) {
		//	if(tc.length<sc.length) {
			//	hantei=false;
		//	}
		//}
*/


		StringBuilder sortS = new StringBuilder();
		StringBuilder sortT = new StringBuilder();
		for(int i=0;i<s.length();i++) {
			sortS.append(sc[i]);
		}
		for(int i=0;i<t.length();i++) {
			sortT.append(tc[i]);
		}
	//	System.out.println(sortS);
	//	System.out.println(sortT);

		String[] zisyo =new String[2];
		zisyo[0]=sortS.toString();
		zisyo[1]=sortT.toString();
		Arrays.sort(zisyo);
	//	System.out.println(zisyo[0]);
	//	System.out.println(zisyo[1]);
boolean hantei = true;
		if(zisyo[0].equals(sortS.toString())) {
			hantei = true;
		}else {
			hantei =false;
		}



		if(hantei==true) {
			if(s.equals(t)) {
				System.out.println("No");
			}else {
			System.out.println("Yes");}
		}else {
			System.out.println("No");
		}


	}

}