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>
int main()
{
	int i,j, jud; // jud =1 yes, =0 no
	char s[101], t[101], letter;
	jud = 0;
	// I scan and rearrange s:
	scanf("%s",s);
	for (j=0;j<=100;j++){
	for (i=0;s[i+1]!=0;i++){
		if (s[i]>s[i+1]){ letter=s[i]; s[i]=s[i+1]; s[i+1]=letter;}
		else ;
	} }
	// next, I do the same for t:
		scanf("%s",t);
	for (j=0;j<=100;j++){
	for (i=0;t[i+1]!=0;i++){
		if (t[i]<t[i+1]){ letter=t[i]; t[i]=t[i+1]; t[i+1]=letter;}
		else ;
	} }
	// printf("%s\n",s); printf("%s\n",t);
	// Lastly, I compare s and t:
	j=0;
	for (i=0;j==0&&s[i]!=0;i++){
		if (s[i]>t[i]) j=1; 
			else{ 	if(s[i]<t[i]){ jud = 1; j=1;} 
					else ; }
	}
	if (jud==0) printf("No"); else printf("Yes");
	return 0;
}