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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<stdio.h>
 
int main()
{
  int i=0,j=0,s_max=0,t_max=0;
  int s_baketu[26]={0};
  int t_baketu[26]={0};
  char s[200]={0},t[200]={0};
/*
  for(i=0;i<100;i++)printf("%d ",s[i]);
  printf("\n");
  for(i=0;i<100;i++)printf("%d ",t[i]);
  printf("\n");
*/  scanf("%s",s);
  scanf("%s",t);
/*  
  for(i=0;i<100;i++)printf("%d ",s[i]);
  printf("\n");
  for(i=0;i<100;i++)printf("%d ",t[i]);
  printf("\n");
*/
  for(i=0;i<200;i++){
    if(s[i]!=0)s_baketu[s[i]-96]++;
  } 
  for(i=0;i<26;i++){
    while(s_baketu[i]>0){
      s[s_max]=i;
      s_baketu[i]--;
      s_max++;
    }      
  }
  
  for(i=0;i<200;i++){
    if(t[i]!=0)t_baketu[t[i]-96]++;
  }
    for(i=0;i<26;i++){
    while(t_baketu[25-i]>0){
      t[t_max]=25-i;
      t_baketu[25-i]--;
      t_max++;
    }      
  }
/*  
  for(i=0;i<3;i++)printf("%d ",s[i]);
  printf("\n");
  for(i=0;i<3;i++)printf("%d ",t[i]);
  printf("\n");
*/ 
  for(i=0;i<200&&j==0;i++){
    if(s[i]<t[i])j=1;
    if(s[i]>t[i]&&j==0){
     printf("No");
     return 0;
   }
  }
  if (j==1)printf("Yes");
  else if(j==0)printf("No");
  return 0;
}