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: Wrong Answer)

#include<stdio.h>

int len(char a[]){
  int i=0;

  while(a[i]!=0) i++;

  return i;
}

void sort(char a[],int n){
  int i,j;
  char c;

  for(i=1;i<n;i++){
    for(j=n-1;j>i;j--){
      if(a[j]<a[j-1]){
        c = a[j];
        a[j] = a[j-1];
        a[j-1] = c;
      }
    }
  }

}

void reverse(char a[],int n){
  int i;
  char b;
  for(i=0;i<n/2;i++){
    b = a[i];
    a[i] = a[n-i-1];
    a[n-i-1] = b;
  }
}

int main(void){
  char a[1000];
  char b[1000];
  int i,n,m,c=0;

  scanf("%s %s",a,b);

  n = len(a);
  m = len(b);
  sort(a,n);
  sort(b,m);
  reverse(b,m);

  for(i=0;a[i]!=0 && b[i]!=0;i++){
    if(b[i]==a[i]) c++;
    if(b[i] > a[i]) break;
  }


  if(a[i]!=0 || n==c && n!=m) printf("Yes\n");
  else printf("No\n");

  return 0;
}