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>
#include<stdlib.h>
#include<string.h>

int main() {
	//変数宣言
	int i, j;
	int flag;
	char Sin[200002], Tin[200002];
	char c;
	
	//入力受付
	scanf("%s", Sin);
	scanf("%s", Tin);
	
	int size = strlen(Sin);

	for (i = 0; i < size; i++) {
		if (Sin[i] != Tin[i]) {
			c = Sin[i];
			for (j = 0; j < size; j++) {
				if (Sin[j] == c || Sin[j] == Tin[i]) {
					if (Sin[j] == c) Sin[j] = Tin[j];
					else if(Sin[j] == Tin[i]) Sin[j] = c;
					if (Sin[j] != Tin[j]) {
						printf("No\n");
						return 0;
					}
				}
			}
			Sin[i] = c;
		}
	}
	printf(Sin == Tin ? "Yes\n":"No\n");
	return 0;
}