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 Java7 (OpenJDK 1.7.0) (judge verdict: Accepted)

 

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

 class Main {
    public static void main(String[]args){
        Scanner scan = new Scanner(System.in);
        char s1[] = scan.next().toCharArray();
        char s2[] = scan.next ().toCharArray();
        Character[] newarr = new Character[s2.length];
        int i =0 ;
        for(char ch : s2){
            newarr[i++] = Character.valueOf(ch) ; 
        }
        Arrays.sort(s1 );
        Arrays.sort( newarr,Collections.reverseOrder());
        StringBuilder sb = new StringBuilder();
        for(Character x : newarr){
            sb.append(x);
        }
         
        //System.out.println(String.valueOf(s1)  + " "+sb.toString());
        if(String.valueOf(s1).compareTo(sb.toString()) < 0){
            System.out.println("Yes");
        }
        else{
            System.out.println("No");
        }
        
        
    }
    
    
}