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.*;
import java.io.*;

public class Main{
  public static void main(String args[]) throws Exception{
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    String s[] = br.readLine().split("");
    String t[] = br.readLine().split("");
    Arrays.sort(s);
    Arrays.sort(t,Comparator.reverseOrder());
    int flag = 0;
    if(s.length < t.length){
      for(int i = 0;i < s.length - 1;i++){
        if(s[i].equals(t[i])){
          if(s[i+1].compareTo(t[i+1]) < 0){
            flag = 1;
            break;
          }
        }
      }
    }else{
      for(int i = 0;i < t.length - 1;i++){
        if(s[i].equals(t[i])){
          if(s[i+1].compareTo(t[i+1]) < 0){
            flag = 1;
            break;
          }
        }
      }
    }
    if(s[0].compareTo(t[0]) <= 0){
      flag = 1;
    }
    if(flag == 1){
      System.out.println("Yes");
    }else{
      System.out.println("No");
    }
  }
}