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 Java7 (OpenJDK 1.7.0) (judge verdict: Accepted)

import java.util.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String s = sc.next(), t = sc.next();
		String ans[] = { "Yes", "No" };
		sc.close();
		char s2[] = s.toCharArray();
		char t2[] = t.toCharArray();

		Arrays.sort(s2);
		Arrays.sort(t2);
		int n = t2.length;
		for (int i = 0; i < n / 2; ++i) {
			char tmp = t2[n - i - 1];
			t2[n - i - 1] = t2[i];
			t2[i] = tmp;
		}
		int a = 0;
		int n2 = s2.length;
		for (int i = 0; i < n2; ++i) {
			if (s2[i] > t2[i]) {
				a = 1;
				break;
			} else if (s2[i] < t2[i])
				break;
			if (i > n - 2) {
				a = 1;
				break;
			}
		}
		System.out.println(ans[a]);
	}

}