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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>

void swap(char* pX, char* pY)
{
    char tmp;

    tmp = *pX;
    *pX = *pY;
    *pY = tmp;
}

int main(void)
{
    char a[101], b[101];
    int numA=0;
    int numB=0;

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

    while('\0' != a[numA]) {
        numA++;
    }

    while('\0' != b[numB]) {
        numB++;
    }

    for(int i=0; numA>i; i++) {
        for(int j=numA-1; i<j; j--) {
            if(a[j] < a[j-1]) {
                swap(&a[j], &a[j-1]);
            }
        }
    }

    for(int i=0; numB>i; i++) {
        for(int j=numB-1; i<j; j--) {
            if(b[j] > b[j-1]) {
                swap(&b[j], &b[j-1]);
            }
        }
    }

    if (a < b) {
        printf("Yes\n");
    }
    else {
        printf("No\n");
    }

    /*
    int i=0;
    while(i<=numA) {
        if(a[i] < b[i]) {
            printf("Yes\n");
            return 0;
        }
        else if((a[i] > b[i]) || ('\0'==b[i])) {
            printf("No\n");
            return 0;
        }
        else {
            i++;
            continue;
        }
    }

    printf("Yes\n");
    */

    return 0;
}