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 C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
char a[200000+100];
char b[200000+100];
vector<int>aa[26];
vector<int>bb[26];
int main()
{
	cin>> a>> b;
	for(int i=0;i<strlen(a);i++)
		aa[a[i]-'a'].push_back(i);
	for(int i=0;i<strlen(b);i++)
		bb[b[i]-'a'].push_back(i);
	bool flag=false;
	for(int i=0;i<26;i++)
	{
		for(int k=1;k<bb[i].size();k++)
			 if(a[bb[i][k]]!=a[bb[i][k-1]])
			 {
			 	flag=true;
			 	break;
			 }
		if(flag)
			break;
	}
	for(int i=0;i<26;i++)
	{
		for(int k=1;k<aa[i].size();k++)
			 if(b[aa[i][k]]!=b[aa[i][k-1]])
			 	flag=true;
		if(flag)
			break;
	}
	if(!flag)
		printf("Yes\n");
	else
		printf("No\n");
	return 0;
 }