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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>

int lex_smaller(char* s1, char* s2)
{
	int i;
	for (i = 0; s1[i] != 0 && s2[i] != 0; i++) {
		if (s1[i] < s2[i]) return 1;
		else if (s1[i] > s2[i]) return -1;
	}
	if (s1[i] == s2[i]) return 0;
	else if (s1[i] == 0) return 1;
	else return -1;
}

int main()
{
	char s[101], t[101];
	scanf("%s", s);
	scanf("%s", t);

	int i, j, count[2][26] = {};
	for (i = 0; s[i] != 0; i++) count[0][s[i] - 'a']++;
	for (i = 0; t[i] != 0; i++) count[1][t[i] - 'a']++;
	for (i = 0, j = 0; s[i] != 0; i++) {
		for (; count[0][j] == 0; j++);
		s[i] = 'a' + j;
		count[0][j]--;
	}
	for (i = 0, j = 25; t[i] != 0; i++) {
		for (; count[1][j] == 0; j--);
		t[i] = 'a' + j;
		count[1][j]--;
	}
	
	if (lex_smaller(s, t) == 1) printf("Yes\n");
	else printf("No\n");
	fflush(stdout);
	return 0;
}