#include <bits/stdc++.h>
using namespace std;
int n, dd[30], res, id[1000005][30], th[1000005][30];
string s;
void Nhap() {
  cin >> s;
  n = s.length();
}
void QuickSort(int l, int r, int xx) {
  int i = l, j = r;
  int x = id[xx][(l + r) / 2];
  while (i <= j) {
    while (id[xx][i] > x) i++;
    while (id[xx][j] < x) j--;
    if (i <= j) swap(th[xx][i], th[xx][j]), swap(id[xx][i++], id[xx][j--]);
  }
  if (l < j) QuickSort(l, j, xx);
  if (i < r) QuickSort(i, r, xx);
}
void pre() {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 30; j++) {
      th[i][j] = j;
      id[i][j] = -1;
    }
  for (int i = 0; i < n; i++) {
    id[i][s[i] - 'a'] = i;
    if (i > 0)
      for (int j = 0; j < 30; j++)
        if (j != s[i] - 'a') id[i][j] = id[i - 1][j];
  }
  for (int i = 0; i < n; i++) QuickSort(0, 26, i);
}
void process(int x) {
  int l = 0, siz = 1;
  long long dem = 0;
  for (int i = 0; i < 30; i++) dd[i] = -1;
  if (x == 1) dem++, dd[s[0] - 'a'] = 0;
  for (int r = 1; r < n; r++) {
    if (dd[s[r] - 'a'] < 0) siz++;
    if (siz >= x) {
      dem++;
      if (siz > x) {
        dd[th[r][siz - 1]] = -1;
        siz--;
        l = id[r][siz] + 1;
      }
      dem += (long long)id[r][siz - 1] - l;
    }
    dd[s[r] - 'a'] = r;
  }
  cout << dem << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  Nhap();
  for (int i = 0; i < n; i++) {
    if (dd[s[i] - 'a'] == 0) res++;
    dd[s[i] - 'a'] = 1;
  }
  cout << res << "\n";
  pre();
  for (int i = 1; i <= res; i++) process(i);
  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;
int n, dd[30], res, id[1000005][30], th[1000005][30];
string s;
void Nhap() {
  cin >> s;
  n = s.length();
}
void QuickSort(int l, int r, int xx) {
  int i = l, j = r;
  int x = id[xx][(l + r) / 2];
  while (i <= j) {
    while (id[xx][i] > x) i++;
    while (id[xx][j] < x) j--;
    if (i <= j) swap(th[xx][i], th[xx][j]), swap(id[xx][i++], id[xx][j--]);
  }
  if (l < j) QuickSort(l, j, xx);
  if (i < r) QuickSort(i, r, xx);
}
void pre() {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 30; j++) {
      th[i][j] = j;
      id[i][j] = -1;
    }
  for (int i = 0; i < n; i++) {
    id[i][s[i] - 'a'] = i;
    if (i > 0)
      for (int j = 0; j < 30; j++)
        if (j != s[i] - 'a') id[i][j] = id[i - 1][j];
  }
  for (int i = 0; i < n; i++) QuickSort(0, 26, i);
}
void process(int x) {
  int l = 0, siz = 1;
  long long dem = 0;
  for (int i = 0; i < 30; i++) dd[i] = -1;
  if (x == 1) dem++, dd[s[0] - 'a'] = 0;
  for (int r = 1; r < n; r++) {
    if (dd[s[r] - 'a'] < 0) siz++;
    if (siz >= x) {
      dem++;
      if (siz > x) {
        dd[th[r][siz - 1]] = -1;
        siz--;
        l = id[r][siz] + 1;
      }
      dem += (long long)id[r][siz - 1] - l;
    }
    dd[s[r] - 'a'] = r;
  }
  cout << dem << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  Nhap();
  for (int i = 0; i < n; i++) {
    if (dd[s[i] - 'a'] == 0) res++;
    dd[s[i] - 'a'] = 1;
  }
  cout << res << "\n";
  pre();
  for (int i = 1; i <= res; i++) process(i);
  return 0;
}

```