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)

#include <bits/stdc++.h>
using namespace  std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define ll long long

int main(){
  string s, t;
  cin >> s >> t;
  vector<char> str1(s.size());
  vector<char> str2(t.size());
  rep(i, s.size()){
    str1[i] = s[i];
  }
  rep(i, t.size()){
    str2[i] = t[i];
  }
  sort(str1.begin(), str1.end());
  sort(str2.begin(), str2.end(), greater<char>());
  
  bool judge1 = 0;
  bool judge2 = 0;
  rep(i, str1.size()){
    if (str1[i] == str2[i]) {
      judge1 = 1;
      judge2 = 1;
    }
    else if (str1[i] < str2[i]) {
      judge1 = 1;
      break;
    }
    else {
      judge1 = 0;
      break;
    }
  }
  
  if (s == t) judge1 = 0;
  if (judge1 && judge2 && s.size() >= t.size()) judge1 = 0;
  
  if (judge1) cout << "Yes" << endl;
  else cout << "No" << endl;
      
  
}