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<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cmath>
using namespace std;
#define MODE 1
#ifdef MODE
#define DEB(X) cout<< #X <<": "<<X<<" ";
#define ARDEB(i,X) cout<< #X <<"["<<i<<"]: "<<X[i]<<" ";
#define END cout<<"\n";
#else
#define DEB(X) {}
#define ARDEB(i,X) {}
#define END {}
#endif
#define REP(i,n) for(int i=0;i<(n);i++)
#define FOR(i,X,n) for(int i=(X);i<(n);i++)
#define EREP(i,n) for(int i=1;i<=(n);i++)
#define Endl endl
int main(){
	string s,t;
	cin>>s>>t;
	sort(s.begin(),s.end());
	sort(t.begin(),t.end());
	if(s.size()<t.size()){
		cout<<"Yes"<<endl;
		return 0;
	}
	for(int i=0;i<t.size();i++){
		if(t[i]>s[0]){
			cout<<"Yes"<<Endl;
			return 0;
		}
	}
	cout<<"No"<<Endl;
	return 0;
}