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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int num(){
    char s[11];
    int i, n;
    for(i=0; i<sizeof(s); i++){
        s[i] = getchar();
        if(s[i] == ' ' || s[i] == '\n'){
            s[i] = '\0';
            break;
        }
    }
    sscanf(s, "%d", &n);
    return n;
}
int comp(char a[], char b[]){
    int i=0;
    while(a[i] != '\0'){
        if(a[i] != b[i]){
            return a[i] - b[i];
        }
        i++;
    }
    return 0;
}
void sort1(char x[], int n){
    int i, j;
    char y;
    for(i=n; i>1; i--){
        for(j=0; j<=i-2; j++){
            if(x[j] > x[j+1]){
                y = x[j];
                x[j] = x[j+1];
                x[j+1] = y;
            }
        }
    }
}
void sort2(char x[], int n){
    int i, j;
    char y;
    for(i=n; i>1; i--){
        for(j=0; j<=i-2; j++){
            if(x[j] < x[j+1]){
                y = x[j];
                x[j] = x[j+1];
                x[j+1] = y;
            }
        }
    }
}
int main(void){
    char s[101];
    char t[101];
    int i;
    int sl=0, tl=0;
    
    while(s[i] != '\0'){
        sl++;
        i++;
    }
    i = 0;
    while(t[i] != '\0'){
        tl++;
        i++;
    }
    
    sort1(s, sl);
    sort2(t, tl);
    
    if(comp(s, t) < 0){
        printf("Yes\n");
    }
    else{
        printf("No\n");
    }
    
    return 0;
}