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++14 (GCC 5.4.1) (judge verdict: Accepted)

/*
このコード、と～おれ!
Be accepted!
  ∧＿∧　
（｡･ω･｡)つ━☆・*。
⊂　　 ノ 　　　・゜+.
　しーＪ　　　°。+ *´¨)
　　　　　　　　　.· ´¸.·*´¨) ¸.·*¨)
　　　　　　　　　　(¸.·´ (¸.·'* ☆
*/


#define rep(i, n) for(int i = 0; i < (n); ++i)
#define repr(i, n) for(int i = n; i >= 0; --i)
#define printynl(a) printf(a ? "yes\n" : "no\n")
#define printyn(a) printf(a ? "Yes\n" : "No\n")
#define printYN(a) printf(a ? "YES\n" : "NO\n")
#define printfin(a) printf(a ? "possible\n" : "inposible\n")
#define all(x) (x).begin(), (x).end()
using ll = long long;

const int INF = 2147483647;
//const ll LINF = ll();今度書く

//マクロとかここまで

#include <stdio.h>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <vector>

using namespace std;

int main() {

	char tmp[100], tmp2[100];
	bool ispossible;
	scanf("%s%s", tmp, tmp2);
	string s = tmp, t = tmp2;
	sort(all(s));
	sort(all(t), greater<int>());
	rep(i, min(s.length(), t.length())) {
		if (s[i] < t[i]) {
			ispossible = true;
			break;
		}
		else if (s[i] > t[i]) {
			ispossible = false;
			break;
		}
		else {
			if (i == min(s.length(), t.length()) - 1) {
				if (s.length() < t.length()) {
					ispossible = true;
				}
				else {
					ispossible = false;
				}
			}
			continue;
		}
	}
	printyn(ispossible);

	return 0;
}