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>
#include <ctype.h>
#include <math.h>

#define char_max 200000

int Pow(int n, int m);
int Max(int N[], int n);
int Min(int N[], int n);
int Sum(int N[], int n);





int main()
{
	char* S = (char*)calloc(char_max + 1, sizeof(char));
	char* T = (char*)calloc(char_max + 1, sizeof(char));

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

	/*struct Alpha {
		char alpha;
		char ret;
	};

	struct Alpha Alpha_[27] = { {'a',NULL},{'b',NULL},{'c',NULL},{'d',NULL},{'e',NULL},{'f',NULL},{'g',NULL},{'h',NULL},{'i',NULL},{'j',NULL},{'k',NULL},{'l',NULL},{'m',NULL},{'n',NULL},{'o',NULL},{'p',NULL},{'q',NULL},{'r',NULL},{'s',NULL},{'t',NULL},{'u',NULL},{'v',NULL},{'w',NULL},{'x',NULL},{'y',NULL},{'z',NULL} };
	*/

	char list[27] = { NULL };

	for (int i = 0; i < char_max; i++) {
		if (S[i] != T[i]) {
			if (list[(int)T[i] - (int)'a'] != NULL) {
				S[i] = list[(int)list[(int)T[i] - (int)'a']-(int)'a'];
			}else if (list[(int)S[i]-(int)'a'] == NULL) {
				list[(int)S[i] - (int)'a'] = T[i];
				list[(int)T[i] - (int)'a'] = S[i];
				S[i] = list[(int)S[i] - (int)'a'];
			}
			else {
				S[i] = list[(int)S[i] - (int)'a'];
			}
		}
	}

	for (int i = 0; i < char_max; i++) {
		if (S[i] != T[i]) {
			printf("No");
			return 0;
		}
	}

	printf("Yes");
	return 0;

}

int Pow(int n, int m) {
	int ret = 1;

	for (int i = 0; i < m; i++) {
		ret *= n;
	}

	return ret;
}

int Max(int N[], int n) {
	int max = N[0];

	for (int i = 0; i < n; i++) {
		max = max < N[i] ? N[i]: max;
	}

	return max;
}

int Min(int N[], int n) {
	int min = N[0];

	for (int i = 0; i < n; i++) {
		min = min > N[i] ? N[i] : min;
	}

	return min;
}

int Sum(int N[], int n) {
	int sum = 0;

	for (int i = 0; i < n; i++) {
		sum += N[i];
	}

	return sum;
}