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: Wrong Answer)

import java.util.*;
public class Main {
    @SuppressWarnings("unchecked")
    public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
        String s1 = sc.next();
        String[] s1a = new String[s1.length()];
        String s2 = sc.next();
        String[] s2a = new String[s2.length()];
        
        for(int nLp = 0; nLp < s1.length();nLp++) {
            s1a[nLp] = s1.substring(nLp, nLp+1);
        }
        for(int nLp = 0; nLp < s2.length();nLp++) {
            s2a[nLp] = s2.substring(nLp, nLp+1);
        }
//        for(String s : s1a)
//        System.out.println(s);
//        for(String s : s2a)
//        System.out.println(s);
        
        Arrays.sort(s1a,new Comparator() {
            @Override
            public int compare(Object arg0, Object arg1) {
                return arg1.hashCode() - arg0.hashCode();
            }
            
        });        
        Arrays.sort(s2a,new Comparator() {
            @Override
            public int compare(Object arg0, Object arg1) {
                return arg1.hashCode() - arg0.hashCode();
            }
            
        });
//
//        for(String s : s1a)
//        System.out.println(s);
//        for(String s : s2a)
//        System.out.println(s);
        
        StringBuffer sb1 = new StringBuffer();
        for(int nLp = 0; nLp < s1.length();nLp++) {
           sb1.append(s1a[nLp]);
        }
        StringBuffer sb2 = new StringBuffer();
        for(int nLp = 0; nLp < s2.length();nLp++) {
           sb2.append(s2a[nLp]);
        }

//        System.out.println(sb1.toString());
//        System.out.println(sb2.toString());
        // 出力v
        if (sb1.toString().compareTo(sb2.toString()) >= 0) {
            System.out.println("No");
        } else {
            System.out.println("Yes");
        }
}
}