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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int main()
{
	char s[200010], t[200010];
	char f[30], g[30];
	for (int i = 0; i < 30; i++)
	{
		f[i] = '0';
		g[i] = '0';
    }
	scanf("%s\n%s", &s, &t);
	for (int i = 0; s[i] != '\0'; i++)
	{
		if (f[(int)(s[i] - 'a')] == '0')
		{
			f[(int)(s[i] - 'a')] = t[i];
		}
		else if (f[(int)(s[i] - 'a')] != t[i])
		{
			printf("No\n%d 1",i);
			return 0;
		}
	}
	for (int i = 0; t[i] != '\0'; i++)
	{
		if (g[(int)(t[i] - 'a')] == '0')
		{
			g[(int)(t[i] - 'a')] = s[i];
		}
		else if (g[(int)(t[i] - 'a')] != s[i])
		{
			printf("No\n%d 2",i);
			return 0;
		}
	}
	printf("Yes\n");
	return 0;
}