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)

//わかりません
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

int compDown(const void*a, const void*b){
    return *(char*)b - *(char*)a;
}
//文字列に含まれる文字の種類を数える
int count(char* ps){
    char tmp = *ps;
    int alphaCnt = 1;
    while(*ps != '\0'){
        if(tmp == *ps){ps++;}
        else{
            alphaCnt++;
            tmp = *ps;
            ps++;
        }
    }
    return alphaCnt;
}
//文字列で同じ文字が連続している添字を見つける→できなかった
void continuous(char *ps, int *pd){
    for(int i = 0; i < strlen(ps)-1; i++){
        if(ps[i] == ps[i+1]){pd[i] = 0;}
        else{pd[i] = -1;}
    }
    //?????
}

int main(void){
    char s[200001];
    char t[200001];
    scanf("%s", s);
    scanf("%s", t);

    char c[200001];
    strcpy(c, s);
    qsort(c, 200001, sizeof(char), compDown);
    int sc = count(c);
    strcpy(c, t);
    qsort(c, 200001, sizeof(char), compDown);
    int tc = count(c);

    //文字列に含まれる文字の種類の数が同じ時は
    //tで連続してるところは同じ添字でsも連続していないといけなさそう
    int ans;
    //これはくさい。たとえば、
    //s:abccd t:aaabbみたいに、tで文字の種類が変わる時にsでは変わってなかった時
    //一致できなさそうな気がする
    if(sc > tc){ans = 1;}
    //これはあってそう
    else if(sc < tc){ans = 0;}
    //ここは実装できなかった
    else{
        int d[200001];
        continuous(s, d);
    }

    if(ans == 1){printf("Yes\n");}
    else{printf("No\n");}
    return 0;
}