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[101];
	char t[101];
	int soe;
	int i = 1;
	char tmp;
	int hanbetsu = 0;
	
	scanf("%s", s);
	scanf("%s", t);
	
	while (i != 0){
		i = 0;
		for (soe = 1; s[soe] != '\0'; soe++){
			if (s[soe] < s[soe - 1]){
				tmp = s[soe];
				s[soe] = s[soe - 1];
				s[soe - 1] = tmp;
				i++;
			}
			
		}
		
	}
	
	i = 1;
	
	while (i != 0){
		i = 0;
		for (soe = 1; t[soe] != '\0'; soe++){
			if (t[soe - 1] > t[soe]){
				tmp = t[soe];
				t[soe] = t[soe - 1];
				t[soe - 1] = tmp;
				i++;
			}
		}
	}
	
	
	for (soe = 0; s[soe] != '\0'; soe++){
		if (s[soe] > t[soe]){
			hanbetsu++;
		}
	}
	
	if (hanbetsu != 0){
		printf("\n");
	}
	else {
		printf("No\n");
	}
	
	return (0);
}