#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:36777216")
using namespace std;
char ch[333333];
int flag[27], pos[300001][27];
int n, ds;
long long calc(int x) {
  long long res = 0;
  int tail = 0, sum = 0;
  memset(flag, 0, sizeof(flag));
  for (int i = 0; i < n; i++) {
    while (tail < n) {
      if (sum >= x) break;
      if (flag[ch[tail] - 'a'] == 0) sum++;
      flag[ch[tail] - 'a']++;
      tail++;
    }
    if (tail == n && sum < x) break;
    if (tail == n)
      res++;
    else {
      int mn = n;
      for (int j = 0; j < 26; j++)
        if (flag[j] == 0) mn = min(mn, pos[tail][j]);
      res += (long long)(mn - tail + 1);
    }
    flag[ch[i] - 'a']--;
    if (flag[ch[i] - 'a'] == 0) sum--;
  }
  return res;
}
int main() {
  scanf(" %s", ch);
  n = strlen(ch);
  for (int i = 0; i < n; i++)
    if (!flag[ch[i] - 'a']) flag[ch[i] - 'a'] = 1, ds++;
  for (int i = 0; i < 26; i++) pos[n][i] = n;
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 0; j < 26; j++)
      pos[i][j] = (ch[i] - 'a' == j) ? i : pos[i + 1][j];
  }
  printf("%d\n", ds);
  for (int i = 1; i <= ds; i++) printf("%I64d\n", calc(i));
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
#pragma comment(linker, "/STACK:36777216")
using namespace std;
char ch[333333];
int flag[27], pos[300001][27];
int n, ds;
long long calc(int x) {
  long long res = 0;
  int tail = 0, sum = 0;
  memset(flag, 0, sizeof(flag));
  for (int i = 0; i < n; i++) {
    while (tail < n) {
      if (sum >= x) break;
      if (flag[ch[tail] - 'a'] == 0) sum++;
      flag[ch[tail] - 'a']++;
      tail++;
    }
    if (tail == n && sum < x) break;
    if (tail == n)
      res++;
    else {
      int mn = n;
      for (int j = 0; j < 26; j++)
        if (flag[j] == 0) mn = min(mn, pos[tail][j]);
      res += (long long)(mn - tail + 1);
    }
    flag[ch[i] - 'a']--;
    if (flag[ch[i] - 'a'] == 0) sum--;
  }
  return res;
}
int main() {
  scanf(" %s", ch);
  n = strlen(ch);
  for (int i = 0; i < n; i++)
    if (!flag[ch[i] - 'a']) flag[ch[i] - 'a'] = 1, ds++;
  for (int i = 0; i < 26; i++) pos[n][i] = n;
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 0; j < 26; j++)
      pos[i][j] = (ch[i] - 'a' == j) ? i : pos[i + 1][j];
  }
  printf("%d\n", ds);
  for (int i = 1; i <= ds; i++) printf("%I64d\n", calc(i));
}

```