#include <bits/stdc++.h>
using namespace std;
string s;
int n;
int cant[300001][26];
int id2[300001][26];
int f() {
  vector<bool> ok(26, 0);
  for (int i = 0; i < n; i++) ok[s[i] - 'a'] = 1;
  int cont = 0;
  for (int i = 0; i < 26; i++) cont += ok[i];
  return cont;
}
int main() {
  while (cin >> s) {
    n = s.size();
    memset(id2, -1, sizeof(id2));
    for (int i = n - 2; i >= 0; i--) {
      for (int j = 0; j < 26; j++) id2[i][j] = id2[i + 1][j];
      id2[i][s[i + 1] - 'a'] = i + 1;
    }
    int total = f();
    cout << total << endl;
    long long dev[total + 1];
    memset(dev, 0, sizeof(dev));
    for (int i = 0; i < n; i++) {
      if (i > 0) {
        for (int j = 0; j < 26; j++) cant[i][j] = cant[i - 1][j];
      }
      cant[i][s[i] - 'a']++;
    }
    for (int i = 0; i < n; i++) {
      int mask = 1 << (s[i] - 'a');
      int tam = 1;
      int ultimo = i;
      while (true) {
        int minimo = 1 << 20;
        int letra = -1;
        for (int j = 0; j < 26; j++) {
          if (id2[i][j] != -1 && (mask & (1 << j)) == 0) {
            if (id2[i][j] < minimo) {
              minimo = id2[i][j];
              letra = j;
            }
          }
        }
        if (minimo != (1 << 20)) {
          dev[tam++] += minimo - ultimo;
          mask |= 1 << letra;
          ultimo = minimo;
        } else {
          dev[tam] += n - ultimo;
          break;
        }
      }
    }
    for (int i = 1; i <= total; i++) cout << dev[i] << endl;
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to 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>
using namespace std;
string s;
int n;
int cant[300001][26];
int id2[300001][26];
int f() {
  vector<bool> ok(26, 0);
  for (int i = 0; i < n; i++) ok[s[i] - 'a'] = 1;
  int cont = 0;
  for (int i = 0; i < 26; i++) cont += ok[i];
  return cont;
}
int main() {
  while (cin >> s) {
    n = s.size();
    memset(id2, -1, sizeof(id2));
    for (int i = n - 2; i >= 0; i--) {
      for (int j = 0; j < 26; j++) id2[i][j] = id2[i + 1][j];
      id2[i][s[i + 1] - 'a'] = i + 1;
    }
    int total = f();
    cout << total << endl;
    long long dev[total + 1];
    memset(dev, 0, sizeof(dev));
    for (int i = 0; i < n; i++) {
      if (i > 0) {
        for (int j = 0; j < 26; j++) cant[i][j] = cant[i - 1][j];
      }
      cant[i][s[i] - 'a']++;
    }
    for (int i = 0; i < n; i++) {
      int mask = 1 << (s[i] - 'a');
      int tam = 1;
      int ultimo = i;
      while (true) {
        int minimo = 1 << 20;
        int letra = -1;
        for (int j = 0; j < 26; j++) {
          if (id2[i][j] != -1 && (mask & (1 << j)) == 0) {
            if (id2[i][j] < minimo) {
              minimo = id2[i][j];
              letra = j;
            }
          }
        }
        if (minimo != (1 << 20)) {
          dev[tam++] += minimo - ultimo;
          mask |= 1 << letra;
          ultimo = minimo;
        } else {
          dev[tam] += n - ultimo;
          break;
        }
      }
    }
    for (int i = 1; i <= total; i++) cout << dev[i] << endl;
  }
  return 0;
}

```