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>

int compare(char *, char *, int);

int main(){
    int length, i;
    char A[200001], B[200001];

    scanf("%s %s", A, B);

    //配列の長さを取得
    //while(A[length++] != '\0'); 初期値 length = -1
    length = sizeof(A) / sizeof(A[0]);

    if(compare(A,B,length)) printf("No");
    else printf("Yes");

    return 0;
}

//判定関数
int compare(char *A, char *B, int length){
    //a~zまでの探索と全ての配列要素の探索
    for(int alpha = 97;alpha < 123 ; alpha++){
        int charDiff = 100;
        int saver = 100;
        for(int i=-1; i<length; i++){
            if(alpha==B[i]){
                //異なる差を持つ配列の組が存在したら検出する
                charDiff = abs(A[i] - B[i]);
                if(charDiff<100 && saver !=100 && saver != charDiff){
                    return 1;
                }
                saver = charDiff;
            }
        }
    }
    return 0;
}