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[100],t[100],tmp1[100],tmp2[100];
 int N,M,i,j,r,count=0;
 scanf("%s",s);
 scanf("%s",t);
 for(i=0;s[i]!='\0';i++){
  N++;
 }
 for(i=0;t[i]!='\0';i++){
  M++;
 }
 for(i=0;i<N;i++){
  for(j=i+1;j<N;j++){
   if(strcmp(&s[i],&s[j])>0){
    strcpy(&s[i],&s[j]);
    }
  }
 }
 for(i=0;i<M;i++){
  for(j=i+1;j<M;j++){
   if(strcmp(&t[i],&t[j])>0){
    strcpy(&t[i],&t[j]);
    }
  }
 }
 if(N<M){ 
 for(i=0;i<N;i++){
 if(strcmp(&s[i],&t[i])!=0) count++;
 }
 if(count==0) {
printf("Yes");
 return 0;
}
}
 count=0;
 for(i=0;strcmp(&s[i],&t[i])==0;i++){
 count++;
 }
if(strcmp(&s[count+1],&t[count]+1)<0){
 printf("Yes");
 return 0;
} 
 printf("No");
 return 0;
}