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(void)
{
	char s[110] = {0};
	char t[110] = {0};
	int s_counter = 0;
	int t_counter = 0;
	int s_frag = 0;
	int t_frag = 0;

	int max_s = 0;
	int max_t = 0;
	fflush(stdout);
	scanf("%s", s);
	scanf("%s", t);

	for (int i = 0; i < 110; i++)
	{
		if (s[i] != 0)
		{
			s_counter++;
		}
		if (max_s < s[i])
		{
			max_s = s[i];
			s_frag++;
		}
	}
	for (int i = 0; i < 110; i++)
	{
		if (t[i] != 0)
		{
			t_counter++;
		}
		if (max_t < t[i])
		{
			max_t = t[i];
			t_frag++;
		}
	}

	while (1)
	{


		if ((s_counter == t_counter) && (s_frag == 1) && (t_frag == 1))
		{
			printf("No");
			break;
		}
			if (max_s <= max_t)
			{
				printf("Yes");
				break;
			}
			else
			{
				printf("No");
				break;
			}
		

		
	}
	return 0;
}