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;

using ll = long long;
#define ld long double
#define ull unsigned long long int

#define rep(i, n) for (int i = 0; i < (int)(n); i++)  // (i, 10) i=0～i=9まで
#define repr(i, n) for (int i = n; i >= 0; i--)   // (i, 10) i=10～i=0まで
#define FOR(i, m, n) for (int i = m; i < n; i++)  // (i, 3, 10) i=3～i=9まで
#define SORT(v, n) sort(v, v + n);
#define VSORT(v) sort(v.begin(), v.end());
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) ((int)(x).size())
#define pb push_back
#define ret return

typedef pair<int, int>    pii;
typedef vector<int>       vi;
typedef vector<vi>        vii;
typedef vector<long long> vl;

// 総数を1000000007で割った余り
const long long mod = 1e9 + 7;
#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))

int main() {
    // cin.tie(0);
    // ios::sync_with_stdio(false);
    // cout << fixed << setprecision(5);

    // 入力
    string s;
    cin >> s;
    string t;
    cin >> t;

    // 処理
    sort(all(s));
    sort(all(t), greater<char>());

    // 出力
    cout << (s < t ? "Yes" : "No") << endl;

    return 0;
}