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: Accepted)

#include<stdio.h>
#include<string.h>

#define MYSTRINGMAX 200000 // 配列20万分取る
#define SIZE MYSTRINGMAX + 2 // MYSTRINGMAX+2した値をSIZEに入れたものを定義する SIZE = 200002

int main(void) {
	char s[SIZE], t[SIZE], c1, c2;

	fgets(s, SIZE, stdin);
	fgets(t, SIZE, stdin);

	int length = strlen(s); // lengthに文字列の長さを返すstrlen関数を配列sにて判定する
	
	for (int i = 0; i < length; i++) {
		if (s[i] != t[i]) { // s[i]とt[j]が等しくない場合に操作開始
			c1 = s[i]; // c1にif文で引っかかった文字sを代入
			c2 = t[i]; // c2にif文で引っかかった文字tを代入
			for (int j = 0; j < i; j++) { // 変更し終わった部分を変えてしまう場合は"No"
				if (s[j] == c2 || s[j] == c1) {
					printf("No\n");
					return 0;
				}
			}
			for (int j = i; j < length; j++) {
				if (s[j] == c1) {
					s[j] = c2; // s[j]の値と入っているc2の値を交換する
				}
				else if (s[j] == c2) {
					s[j] = c1; 
				}
			}
		}
	}
	printf("Yes\n"); // それ以外ならYesを実行
	return 0;
}