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>
#include<string.h>

int main(void){
    // s: 入力文字列(操作する方)
    // t: 入力文字列(比較される方)
    // c1: "操作"のためにsの文字を入れる
    // c2: "操作"のためにtの文字を入れる
    // flag: 探索を終了する判定のフラッグ
    char s[200001], t[200001], c1, c2;
    int flag = 0;
    scanf("%s\n%s", s, t);
    int length = strlen(s);
    
    for(int i = 0; i < length; i++){ // tとsが同じか頭から1文字ずつ見ていく
        if(t[i] != s[i]){ // 違う文字があったら"操作"を始める
            // 入れ替える文字を代入しておく
            c1 = s[i];
            c2 = t[i];
            for(int j = 0; j < length; j++){ // sに入れ替える文字があるか頭から探す
                if(s[j] == c1){ // sの中に入れ替える文字があったら
                    if(j < i){
                        // tの位置より前を入れ替えようとした時
                        // 最後の比較は同じにならないのでbreakして早々に処理を終える
                        // forを早く抜けないと実行時間制限超過になる
                        flag = 1;
                        break;
                    }
                    s[j] = c2;
                    // 入れ替え終わった文字を次のifの判定で拾ってしまうので
                    // continueで次のループへ飛ぶ
                    continue;
                }
                if(s[j] == c2){ // 同上
                    if(j < i){
                        flag = 1;
                        break;
                    }
                    s[j] = c1;
                }
            }
        }
        if(flag){
            break;
        }
    }

    if(strcmp(s, t) == 0){
        printf("Yes\n");
    }
    else{
        printf("No\n");
    }
    return 0;
}