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 int nmax = 3e5;
const int omega = 30;
long long ans[nmax + 1];
int lst[omega + 1];
int main() {
  cin.sync_with_stdio(false);
  cin.tie();
  cout.tie();
  string s;
  cin >> s;
  int n = (int)s.size();
  for (int i = 0; i < omega; ++i) lst[i] = -1;
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    lst[s[i] - 'a'] = i;
    vector<int> v;
    for (int j = 0; j < 26; ++j) {
      if (lst[j] != -1) v.push_back(lst[j]);
    }
    v.push_back(-1);
    sort(v.begin(), v.end());
    reverse(v.begin(), v.end());
    cnt = (int)v.size() - 1;
    for (int j = 1; j < (int)v.size(); ++j) {
      ans[j] += (v[j - 1] - v[j]);
    }
  }
  cout << cnt << "\n";
  for (int i = 1; i <= 26 && ans[i] != 0; ++i) {
    cout << ans[i] << "\n";
  }
  return 0;
}