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<algorithm>//sort(all(変数),greater<型名>())　で降順に
#include<bits/stdc++.h>
#include<cmath>//切り上げceil(値)
#include<deque>
#include<iomanip>//setprecision(数字)
#include<iostream>
#include<list>
#include<map>
#include<numeric> //xとyの最大公約数は__gcd(x,y),xとyの最小公倍数は(x*y)/__gcd
#include<queue>//priority_queue<型名> 変数：優先度付きキュー　最大値の探索が早い
#include<set>
#include<stack>
#include<string>
#include<vector>
#define rep0(i,n)  for(ll i=0; i<n; i++)
#define rrep0(i,n) for(ll i=n-1; 0<=i; i--)
#define rep1(i,n)  for(ll i=1; i<=n; i++)
#define rrep1(i,n) for(ll i=n; 1<=i; i--)
#define vll vector<ll>
#define vi vector<int>
#define all(x) (x).begin(),(x).end()
#define INF 1e18
using ll = long long;
using namespace std;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

int main(){
  string s,t;  cin >> s >> t;
  sort(all(s));
  sort(all(t));
  rep0(i,s.size()){
    if(s[i] > t[i]){
      cout << "No";
      return 0;
    }
  }
  cout << "Yes";
}