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 (Clang 3.8.0) (judge verdict: Runtime Error)

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

#define NUM_ALPHABET (26)
static int aiCountS[NUM_ALPHABET] = {0};
static int aiCountT[NUM_ALPHABET] = {0};
static char strS[2*10^5+1];
static char strT[2*10^5+1];

int main()
{
	scanf("%s", strS);
	scanf("%s", strT);

	int L = strlen(strS);
	if (strlen(strT) != L) {
		printf("No");
		return 0;
	}
	if (sizeof(strS) <= L) {
		printf("No");
		return 0;
	}

	//アルファベット文字をカウント
	int i, j;
	for (i = 0; i < L; i++) {
		int iCode;

		iCode = strS[i] - 'a';
		if ((iCode < 0) || (NUM_ALPHABET <= iCode)) {
			printf("No");
			return 0;
		}
		aiCountS[iCode]++;

		iCode = strT[i] - 'a';
		if ((iCode < 0) || (NUM_ALPHABET <= iCode)) {
			printf("No");
			return 0;
		}
		aiCountT[iCode]++;
	}

	//照合。一致したら消していく(0にする)。一致するものがなければNo。
	for (i = 0; i < NUM_ALPHABET; i++) {
		int iNum;
		bool bIsFound = false;

		iNum = aiCountS[i];
		if (0 == iNum) {
			continue;
		}

		for (j = 0; j < NUM_ALPHABET; j++) {
			if (iNum == aiCountT[j]) {
				aiCountT[j] = 0;
				bIsFound = true;
				break;
			}
		}

		if (!bIsFound) {
			printf("No");
			return 0;
		}
	}

	printf("Yes");
	return 0;
}