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: Runtime Error)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main()
{
    char* s;
    char *t;
    int i,j,k;
   int strNum;
   int sMoji[26]={0},tMoji[26]={0};
   
   char a = 'a';
   
    s =(char *)malloc(sizeof(char) * (2*10^5));
    t = (char *)malloc(sizeof(char) * (2*10^5));
    
    scanf("%s",s);
    scanf("%s",t);
    strNum = strlen(s);
    
   
    for(i=0;i<strNum;i++){
        
        //文字列を分解する
        for(j=0;j<26;j++){
            if(((int)a + j) == (int)s[i]) {
                    sMoji[j]++;
                    //printf("a - %d %d\n",sMoji[j],j);
            }
            if(((int)a + j) == (int)t[i]) {
                    tMoji[j]++;
                    //printf("b - %d %d\n",tMoji[j],j);
            }
        }
    }
    
    for(i=0;i<26;i++){
        for(j=0;j<26;j++){
            if(sMoji[i] == tMoji[j]){
                tMoji[j] = -1;
                //printf("kekka - %d %d\n",i,j);
                break;
            }
        }
        if(j>=26){
                break;
        }
    }
    
    // 出力
    if(i<26){
        printf("No\n");
    } else {
        printf("Yes\n");
    }

    //free(s);
    //free(t);
    return 0;
}