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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::*;

fn main() {
    let stdin = stdin();
    let mut stdinlock = stdin.lock();
    let mut str1 = String::new();
    let mut str2 = String::new();
    stdinlock.read_line(&mut str1).unwrap();
    stdinlock.read_line(&mut str2).unwrap();
    let mut bytes1 = str1.trim().as_bytes().to_owned();
    let mut bytes2 = str2.trim().as_bytes().to_owned();
    bytes1.sort();
    bytes2.sort();
    bytes2.reverse();
    //println!("{:?}, {:?}", bytes1, bytes2);
    for (i, val2) in bytes2.iter().enumerate() {
        if i >= bytes1.len() {
            println!("Yes");
            return;
        }
        if *val2 > bytes1[i] {
            println!("Yes");
            return;
        }
        if *val2 < bytes1[i] {
            println!("No");
            return;
        }
    }
    println!("No");
}