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: Accepted)

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <time.h>

#define GYOU_MAX 256
void sort_string(char Text[], int size, int sign) {
	char a;

	for (int i = 0; i < size - 1; i++) {
		for (int j = 0; j < size - 1 - i; j++) {
			if (sign > 0) {
				if (Text[j] > Text[j + 1]) {

					a = Text[j];
					Text[j] = Text[j + 1];
					Text[j + 1] = a;
				}
			}
			else {
				if (Text[j] < Text[j + 1]) {

					a = Text[j];
					Text[j] = Text[j + 1];
					Text[j + 1] = a;

				}
			}
		}

	}

}

int main(void){

	char s[101], t[101];
	scanf("%s%s", s, t);
	sort_string(s, strlen(s),1);
	sort_string(t, strlen(t), 0);
	if (strcmp(s, t) < 0) {
		printf("Yes\n");
	}
	else {
		printf("No\n");
	}

	return 0;
}