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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 300005, maxx = 31;
string s;
long long ans[maxx], a[maxx][maxx], p[maxx], f[maxx], x, flag, t;
int main() {
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    f[s[i] - 'a']++;
  }
  for (int i = 0; i < 26; i++) {
    if (f[i]) {
      x++;
      f[i] = 0;
    }
  }
  cout << x << endl;
  t = 0;
  for (int j = 0; j < s.size(); j++) {
    if (!f[s[j] - 'a']) {
      t++;
      f[s[j] - 'a']++;
      p[t] = j;
      for (int k = 0; k < 26; k++) {
        a[t][k] = f[k];
      }
    } else {
      f[s[j] - 'a']++;
    }
  }
  p[x + 1] = s.size();
  for (int i = 1; i <= x; i++) {
    ans[i] += p[i + 1] - p[i];
  }
  for (int i = 1; i < s.size(); i++) {
    for (int j = 1; j <= x; j++) {
      flag = 0;
      a[j][s[i - 1] - 'a']--;
      if (a[j][s[i - 1] - 'a']) {
        continue;
      }
      for (int k = p[j] + 1; k < s.size(); k++) {
        if (!a[j][s[k] - 'a']) {
          a[j][s[k] - 'a']++;
          flag = 1;
          p[j] = k;
          break;
        }
        a[j][s[k] - 'a']++;
      }
      if (!flag) {
        p[j] = s.size();
      }
    }
    for (int j = 1; j <= x; j++) {
      ans[j] += p[j + 1] - p[j];
    }
  }
  for (int i = 1; i <= x; i++) {
    cout << ans[i] << endl;
  }
  return 0;
}