#include <bits/stdc++.h>
void sortLetters(int* inputLetters, int* sortedLetters);
int main(int argc, char* argv[]) {
  char inputString[300003];
  scanf("%s", inputString);
  int length = strlen(inputString);
  long long diversity[29 - 1] = {};
  int i = 0;
  int lastLetterPositions[29] = {};
  while (i <= 26) {
    lastLetterPositions[i] = -1;
    ++i;
  }
  i = 0;
  while (i < length) {
    lastLetterPositions[inputString[i] - 'a'] = i;
    int sortedLetterPositions[29] = {};
    sortLetters(lastLetterPositions, sortedLetterPositions);
    int j = 0;
    for (j = 26; j > 0; --j) {
      diversity[26 - j + 1] +=
          sortedLetterPositions[j] - sortedLetterPositions[j - 1];
    }
    ++i;
  }
  int number = 0;
  i = 1;
  while (i <= 26) {
    if (diversity[i] > 0) {
      ++number;
    }
    ++i;
  }
  printf("%d\n", number);
  i = 1;
  while (i <= 26) {
    if (diversity[i]) {
      printf("%I64d\n", diversity[i]);
    }
    ++i;
  }
  return 0;
}
void sortLetters(int* inputLetters, int* sortedLetters) {
  int i = 0;
  while (i <= 26) {
    sortedLetters[i] = inputLetters[i];
    ++i;
  }
  i = 0;
  while (i <= 26) {
    int j = 0;
    while (j < 26) {
      if (sortedLetters[j] > sortedLetters[j + 1]) {
        int temp = sortedLetters[j + 1];
        sortedLetters[j + 1] = sortedLetters[j];
        sortedLetters[j] = temp;
      }
      ++j;
    }
    ++i;
  }
}

### Prompt

In Cpp, your task is to solve the following problem:

String diversity is the number of symbols that occur in the string at least once. Diversity of s will be denoted by d(s). For example , d("aaa")=1, d("abacaba")=3.

Given a string s, consisting of lowercase Latin letters. Consider all its substrings. Obviously, any substring diversity is a number from 1 to d(s). Find statistics about substrings diversity: for each k from 1 to d(s), find how many substrings of s has a diversity of exactly k.

Input

The input consists of a single line containing s. It contains only lowercase Latin letters, the length of s is from 1 to 3·105.

Output

Print to the first line the value d(s). Print sequence t1, t2, ..., td(s) to the following lines, where ti is the number of substrings of s having diversity of exactly i.

Examples

Input

abca


Output

3
4
3
3


Input

aabacaabbad


Output

4
14
19
28
5

Note

Consider the first example.

We denote by s(i, j) a substring of "abca" with the indices in the segment [i, j].

  * s(1, 1) =  "a", d("a") = 1
  * s(2, 2) =  "b", d("b") = 1
  * s(3, 3) =  "c", d("c") = 1
  * s(4, 4) =  "a", d("a") = 1
  * s(1, 2) =  "ab", d("ab") = 2
  * s(2, 3) =  "bc", d("bc") = 2
  * s(3, 4) =  "ca", d("ca") = 2
  * s(1, 3) =  "abc", d("abc") = 3
  * s(2, 4) =  "bca", d("bca") = 3
  * s(1, 4) =  "abca", d("abca") = 3



Total number of substring with diversity 1 is 4, with diversity 2 equals 3, 3 diversity is 3.

### Response

```cpp
#include <bits/stdc++.h>
void sortLetters(int* inputLetters, int* sortedLetters);
int main(int argc, char* argv[]) {
  char inputString[300003];
  scanf("%s", inputString);
  int length = strlen(inputString);
  long long diversity[29 - 1] = {};
  int i = 0;
  int lastLetterPositions[29] = {};
  while (i <= 26) {
    lastLetterPositions[i] = -1;
    ++i;
  }
  i = 0;
  while (i < length) {
    lastLetterPositions[inputString[i] - 'a'] = i;
    int sortedLetterPositions[29] = {};
    sortLetters(lastLetterPositions, sortedLetterPositions);
    int j = 0;
    for (j = 26; j > 0; --j) {
      diversity[26 - j + 1] +=
          sortedLetterPositions[j] - sortedLetterPositions[j - 1];
    }
    ++i;
  }
  int number = 0;
  i = 1;
  while (i <= 26) {
    if (diversity[i] > 0) {
      ++number;
    }
    ++i;
  }
  printf("%d\n", number);
  i = 1;
  while (i <= 26) {
    if (diversity[i]) {
      printf("%I64d\n", diversity[i]);
    }
    ++i;
  }
  return 0;
}
void sortLetters(int* inputLetters, int* sortedLetters) {
  int i = 0;
  while (i <= 26) {
    sortedLetters[i] = inputLetters[i];
    ++i;
  }
  i = 0;
  while (i <= 26) {
    int j = 0;
    while (j < 26) {
      if (sortedLetters[j] > sortedLetters[j + 1]) {
        int temp = sortedLetters[j + 1];
        sortedLetters[j + 1] = sortedLetters[j];
        sortedLetters[j] = temp;
      }
      ++j;
    }
    ++i;
  }
}

```