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<string.h> 
int main()
{int len_a,len_b;
int cnt=0,cen=0;
	char a[1005],b[1005];
	gets(a);
	len_a=strlen(a);
  gets(b);
  len_b=strlen(b);
	
	if(len_a!=len_b)
	printf("NO\n");
	for(int i=1;i<=len_a;i++)
	{for(int j=i;j<=len_a;j++)
	{
	if(a[i]==a[j])	cnt++;
	}
	}
	
	for(int i=1;i<=len_b;i++)
	{for(int j=i;j<=len_b;j++)
	{if(b[i]==b[j]) cen++;
	}
	}
	if(cen!=cnt)
	printf("NO\n");
	else printf("YES\n");
	return 0;
 }