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 n[101], m[101];
  int i, a, b, temp, j, c;
  scanf("%s", n);
  scanf("%s", m);
  a=strlen(n);
  b=strlen(m);
  for(i=0; i<a; i++){
      for(j=a-1; j>i; j--){
        if(n[j] < n[j-1]){
          temp = n[j];
          n[j] = n[j-1];
          n[j-1] = temp;
        }
      }
    }
    for(i=0; i<b; i++){
        for(j=b-1; j>i; j--){
          if(n[j] < m[j-1]){
            temp = m[j];
            m[j] = m[j-1];
            m[j-1] = temp;
          }
        }
      }
    if(a<b){
      c=a;
    }else{
      c=b;
    }
      for(i=0;i<c;i++){
        if(n[i]<m[i]){
          printf("No\n");
          return 0;
        }else if(n[i]>m[i]){
          printf("Yes\n");
          return 0;
        }else{}
      }
      printf("No\n");
      return 0;
    }