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)

#pragma region
#include <bits/stdc++.h>
using namespace std;
inline void ci(void){
  return;
}
template <typename First, typename... Rest>
void ci(First& first, Rest&... rest){
  cin >> first;
  ci(rest...);
  return;
}
inline void co(void){
  cout << "\n";
  return;
}
template <typename First, typename... Rest>
void co(First first, Rest... rest){
  cout << first << " ";
  co(rest...);
  return;
}
#define ll long long
#define ld long double
#define st string
#define dl double
#define wh while
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define tp tuple
#define ft first
#define sc second
#define ve vector
#define tos to_string
#define len length
#define rt return 0
#define br break
#define con continue
#define lcm(n, m) (n * m / gcd(n, m))
#define np next_permutation
#define repi(i, m, n) for(int i=(int)(m);i<(int)(n);i++)
#define rep(i, n) repi(i, 0, n)
#define repill(i, m, n) for(ll i=(ll)(m);i<(ll)(n);i++)
#define repll(i, n) repill(i, 0, n)
#define revi(i, m, n) for(int i=(int)(m);i>(int)(n);i--)
#define rev(i, n) revi(i, 0, n)
#define revill(i, m, n) for(ll i=(ll)(m);i>(ll)(n);i--)
#define revll(i, n) revill(i, 0, n)
#define sz(v) ((int)(v).size()) 
#define each(i, n) for(auto&& i: n)
#define all(n) (n.begin(), n.end())
#define allr(n) (n.rbegin(), n.rend())
#define sp(n) cout << fixed << setprecision(n)
#pragma endregion
// sqrt ルート, int 2*10^9, gcd　最大公約数
// __builtin_popcount bitの個数

int main() {
  st s, t;
  ci(s, t);
  int N = sz(s), M = sz(t);
  ve<char> S(N), T(M);
  rep(i, N){
    S.at(i) = s.at(i);
  }
  rep(i, M){
    T.at(i) = t.at(i);
  }
  sort all(S);
  sort all(T);
  if(S.at(0) < T.at(M - 1)){
    co("Yes");
    rt;
  }
  else if(N < M){
    bool f = 1;
    rep(i, N){
      if(S.at(i) != T.at(i)){
        f = 0;
      }
    }
    if(f){
      co("Yes");
      rt;
    }
    else{
      co("No");
      rt;
    }
  }
  co("No");
}