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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
#define pb push_back
#define mp make_pair
#define f first
#define s second
#define sc(n) scanf("%d",&n)
#define scl(n) scanf("%lld",&n)
#define pr(n) printf("%d",n)
#define prl(n) printf("%lld",n)
#define nl printf("\n")
#define fast_io ios_base::sync_with_stdio(false);cin.tie(NULL)
ll pow1(ll a, ll b)
{
	ll ans=1;
	while(b)
	{
		if(b&1)
		{
			ans=ans*a;
		}
		a=a*a;
		b=b>>1;
	}
	return ans;
}
set<int> gr[26];
int vis[26],f = 0;
// void dfs(int u,int v)
// {
// 	vis[u] = 1;
// 	if(u == v)
// 	{
// 		f = 1;
// 		return; 
// 	}
// 	vector<int>::iterator it = gr[u].begin();
// 	for(;it!=gr[u].end();it++)
// 	{
// 		if(f == 1)
// 			return;
// 		if(!vis[*it])
// 			dfs(*it,v);
// 	}
// 	if(f == 1)
// 		return;
// }
int main()
{
	string s,t;
	int i,j;
	cin>>s>>t;
	int n = s.size();
	if(s == t)
		cout<<"Yes"<<endl;
	for(i=0;i<n;i++)
		gr[t[i]-'a'].insert(s[i]-'a');
	// for(i=0;i<n;i++)
	// {
	// 	f = 0;
	// 	memset(vis,0,sizeof(vis));
	// 	dfs(s[i]-'a',t[i]-'a');
	// 	if(!f)
	// 	{
	// 		cout<<"No"<<endl;
	// 		return 0;
	// 	}
	// }
	for(i=0;i<26;i++)
		if(gr[i].size()>=2)
		{
			cout<<"No"<<endl;
			return 0;
		}
	cout<<"Yes"<<endl;
	return 0;
}