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>

#define NEW(p,n){p=malloc((n)*sizeof(p[0]));if(p==NULL){printf("not enough memory\n");exit(1);};}
//pの型の変数n個の要素分のメモリを確保し、そのアドレスをpに代入するマクロ

#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))]

//char型のポインタ(=配列と意味は同じ)をString型と定義([0]から数える)
typedef char* String;

//S[0]から始まる文字列の長さを求める関数
int string_len(String str){
    int len=0;
    while(str[len]!=0){
        len++;
    }
    return len;
}

//文字列を標準入力から読み込み、それをString型のメモリを確保し直してくれる関数
//S[0]~S[n-1]までが入力で、S[n]は0になる
//文字列長さの上限は1000まで
String string_input(void){
    int i,len;
    char buf[200001];
    String str;
    scanf("%s",buf);//buf=入力した文字列が入るchar型の配列
    
    len=string_len(buf);//len=文字列の長さを表すint型変数
    NEW(str,len+1);//strはただのポインタだから、ここでNEWを使ってメモリを確保しなくちゃいけない
    for(i=0;i<len;i++){
        str[i]=buf[i];
    }
    str[len]=0;
    
    return str;
}


int main(void){
    String S,T;
    S=string_input();
    T=string_input();

    int N=string_len(S);

    char trans[27];
    for(int i=0;i<26;i++) trans[i]='-';

    for(int i=0;i<N;i++){
        int s=S[i]-'a';
        int t=T[i]-'a';
        if(trans[s]=='-' && trans[t]=='-'){
            trans[s]=T[i];
            trans[t]=S[i];
        }
        else if(trans[s]=='-' && trans[t]!='-'){
            trans[s]=T[i];
            if(trans[t]!=S[i]){
                printf("No\n");
                return 0;
            }
        }
        else if(trans[s]!='-' && trans[t]=='-'){
            trans[t]=S[i];
            if(trans[s]!=T[i]){
                printf("No\n");
                return 0;
            }
        }
        else{
            if(trans[s]!=T[i] || trans[t]!=S[i]){
                printf("No\n");
                return 0;
            }
        }
    }

    //for(int i=0;i<26;i++) printf("%c ",trans[i]);

    printf("Yes\n");
    return 0;
}