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() {
  char S[200001], T[200001];

  long int i, j, k, len, l, tmpl, can_change;

  char c, d, tmpc;

  scanf("%s", S);
  scanf("%s", T);

  l = len = strlen(S);
  k = 0;
  for (i = 0; i < l; i++) {
    c = S[i];
    if (c >= 'a') {
      k++;
      tmpl = i;
      for (j = i; j < l; j++) {
        if ((tmpc = S[j]) == c) {
          S[j] = k;
          tmpl = j + 1;
        } else if (tmpc >= 'a') {
          tmpl = j + 1;
        }
      }
      l = tmpl;
    }
  }
  can_change = k >= 26;

  l = len;
  k = 0;
  for (i = 0; i < l; i++) {
    c = T[i];
    if (c >= 'a') {
      k++;
      tmpl = i;
      for (j = i; j < l; j++) {
        if ((tmpc = T[j]) == c) {
          T[j] = k;
          tmpl = j + 1;
        } else if (tmpc >= 'a') {
          tmpl = j + 1;
        }
      }
      l = tmpl;
    }
  }

  for (i = 0; i < len; i++) {
    c = S[i];
    d = T[i];
    if (c < d) {
      printf("No\n");
      return 0;
    } else if (c != d) {
      if (!can_change) {
        printf("No\n");
        return 0;
      }
      for (j = i; j < len; j++) {
        if ((tmpc = S[j]) == c) {
          S[j] = d;
        }
      }
    }
  }

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