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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)


xa = list(input())

ya = list(input())

an=[]

f=1

allxy=[]
allyx=[]

for i in range(len(xa)):
    if xa[i]!=ya[i]:
        allxy.append([xa[i], ya[i]])
        allyx.append([ya[i], xa[i]])

all2=list(map(list, set(map(tuple, allxy))))
all2.sort()
all3=list(map(list, set(map(tuple, allyx))))
all3.sort()


if len(all2)>1:
    for i in range(len(all2)-1):
        if all2[i][0]==all2[i+1][0]:
            f=-1
            break
        elif all3[i][0]==all3[i+1][0]:
            f=-1
            break
    else:
        if all2[i][0]==all2[i+1][0]:
            f=-1
        elif all3[i][0]==all3[i+1][0]:
            f=-1
else:
    f=1
        
if f>0:
    print("Yes")
else:
    print("No")