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: Wrong Answer)

#include<bits/stdc++.h>
// Begin Header {{{
#define all(x) (x).begin(), (x).end()
#define lli long long int
#define rep(i,n) for(lli i=0; i<n; i++)
#define loop(i, j, n) for(lli i=j; i<n; i++)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
constexpr int INF  = 0x3f3f3f3f;
const long long mod=1e9+7;
//Function
lli combi(lli n, lli r)
{
	if(r==0 || r==n) return 1;
	else return combi(n-1,r) + combi(n-1,r-1);
}
lli sigma1(lli n){return n*(n+1)/2;}
lli sigma2(lli s, lli n){return sigma1(n) - sigma1(s -1);}
using namespace std;
// }}} End Header

int main()
{
	ios::sync_with_stdio(false); cin.tie(nullptr);
	int n, k;
	string s, t;
	cin >> s >> t;
	n = s.size();
	k = t.size();

	int z = Min(n,k);

	vector<char> a(n);
	vector<char> b(k);
	rep(i,n) a[i] = s[i];
	rep(i,k) b[i] = t[i];

	sort(all(a));
	sort(all(b), greater<int>());

	rep(i,n) s[i] = a[i];
	rep(i,k) t[i] = b[i];

	set<string> m;
	m.insert(s);
	m.insert(t);

	auto itr = m.begin();

	if(*itr == s) cout << "Yes" << endl;
	else cout << "No" << endl;
	return 0;
}