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 <string.h>

int main(void){
  char s[101], t[101], tmp;
  int ls, lt, i, j;
  
  scanf("%s %s", s, t);

  ls = strlen(s);
  lt = strlen(t);
 
  for( i = 0; i < ls; i++){
    for( j = i+1; j < ls; j++){
      if(s[i] > s[j]){
	tmp = s[i];
	s[i] = s[j];
	s[j] = tmp;
      }
    }
  }
  for( i = 0; i < lt; i++){
    for( j = i+1; j < lt; j++){
      if(t[i] < t[j]){
	tmp = t[i];
	t[i] = t[j];
	t[j] = tmp;
      }
    }
  }

 
  i = 0;
  while(i < lt){
    if(s[i] < t[i]){
      printf("YES\n");
      break;
    }
    else if(s[i] == t[i]){
      i++;
    }
    else{
      printf("NO\n");
      break;
    }
  }
  if(i == lt)printf("NO");
  return 0;
}