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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include <string.h>

int main(){
  char s[100], t[100];
  int i, j, a, b, tmp;
  scanf("%s", s); a=strlen(s);
  scanf(" %s", t); b=strlen(t);
  
  for(i=0; i<a-1; i++){
    for(j=0; j<a-1-i; j++){
      if(s[j]>s[j+1]){
        tmp=s[j];s[j]=s[j+1];s[j+1]=s[j];
      }
    }
  }

  for(i=0; i<b-1; i++){
    for(j=0; j<b-1-i; j++){
      if(t[j]<t[j+1]){
        tmp=t[j];t[j]=t[j+1];t[j+1]=t[j];
      }
    }
  }

  for(i=0; ; i++){
    if(s[i]=='\0' && t[i]!='\0'){printf("Yes\n"); break;}
    if(t[i]=='\0'){printf("No\n"); break;}
    if(s[i]<t[i]){
      printf("Yes\n"); break;
    }else if(s[i]>t[i]){
      printf("No\n"); break;
    }else{
      continue;
    }
  }

}