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>

int main(void)
{
	char S[200001];
	char T[200001];

	int i = 0,j = 0;
	int flg = 0;
	char keep1,keep2;

	scanf("%s", S);
	scanf("%s", T);


	for (i = 0; S[i] != '\0'; i++) {
		if (S[i] != T[i]) {
			keep1 = S[i];
			keep2 = T[i];
			for (j = 0; S[j] != '\0'; j++) {
				if (keep1 == S[j]) {
					if (j < i) {
						flg = flg - (flg + 1);
						break;
					}
					S[j] = keep2;			
					flg++;
				}
				else if (keep2 == S[j]) {
					if (j < i) {
						flg = flg - (flg + 1);
						break;
					}
					S[j] = keep1;
					flg++;
				}

			}
		}
		if (flg < 0) {
			break;
		}
	
	}

if (flg < 0) {
		printf("No\n");
	}
	else {
		printf("Yes\n");
	}
	
	return 0;
}