Problem:
Score : 300 points

Problem Statement

You are given strings S and T consisting of lowercase English letters.

You can perform the following operation on S any number of times:

Operation: Choose two distinct lowercase English letters c_1 and c_2, then replace every occurrence of c_1 with c_2, and every occurrence of c_2 with c_1.

Determine if S and T can be made equal by performing the operation zero or more times.

Constraints

1 \leq |S| \leq 2 \times 10^5

|S| = |T|

S and T consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

S
T

Output

If S and T can be made equal, print Yes; otherwise, print No.

Sample Input 1

azzel
apple

Sample Output 1

Yes

azzel can be changed to apple, as follows:

Choose e as c_1 and l as c_2. azzel becomes azzle.

Choose z as c_1 and p as c_2. azzle becomes apple.

Sample Input 2

chokudai
redcoder

Sample Output 2

No

No sequences of operation can change chokudai to redcoder.

Sample Input 3

abcdefghijklmnopqrstuvwxyz
ibyhqfrekavclxjstdwgpzmonu

Sample Output 3

Yes

Solution in Python (3.8.2) (judge verdict: Wrong Answer)

import sys

input = sys.stdin.readline

S = input()
S = S.replace('\n','')
s_list = list(S)

T = input()
T = T.replace('\n','')
t_list = list(T)

alpha_dict = {}
t_diff_count = [0 for i in range(27)]
s_diff_count = [0 for i in range(27)]

for i,c in enumerate(range(ord('a'),ord('z')+1)):
    alpha_dict[chr(c)] = i 

for i,s in enumerate(s_list):
    if not s == t_list[i]:
        number = alpha_dict[s_list[i]]
        s_diff_count[number] += 1
        number = alpha_dict[t_list[i]]
        t_diff_count[number] += 1

for i,s in enumerate(s_list):
    number = alpha_dict[s_list[i]]
    s_count = s_diff_count[number] 
    number = alpha_dict[t_list[i]]
    t_count = t_diff_count[number]
    if not s_count == t_count:
        if t_count >= 2:
            print("No")
            exit(0)
print("Yes")