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;
void add(vector<char>& vc, vector<int>& vi, char c, int i) {
  int index = find(vc.begin(), vc.end(), c) - vc.begin();
  if (index != vc.size()) {
    vc.erase(vc.begin() + index);
    vi.erase(vi.begin() + index);
  }
  vc.push_back(c);
  vi.push_back(i);
}
int main() {
  ios_base::sync_with_stdio(false);
  string s;
  cin >> s;
  vector<char> characters;
  vector<int> indexes;
  add(characters, indexes, '#', s.size());
  vector<long long> ans(30, 0);
  for (int i = s.size() - 1; i >= 0; --i) {
    add(characters, indexes, s[i], i);
    for (int j = indexes.size() - 1; j > 0; --j) {
      ans[indexes.size() - j] += indexes[j - 1] - indexes[j];
    }
  }
  cout << indexes.size() - 1 << endl;
  for (int i = 1; ans[i] != 0; ++i) {
    cout << ans[i] << endl;
  }
  return 0;
}