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;
int main(int argc, char* argv[]) {
  string s;
  cin >> s;
  int size = s.size();
  vector<long long int> cnt(26 + 1, 0);
  vector<pair<int, char> > last(26);
  for (int i = 0; i < 26; ++i) last[i] = make_pair(-1, 'a' + i);
  for (int i = 0; i < size; ++i) {
    for (int j = 0; j < 26; ++j) {
      if (last[j].second == s[i]) {
        last[j].first = i;
        break;
      }
    }
    sort(last.begin(), last.end());
    int hi = i;
    int k = 1;
    for (int j = 24; j >= 0; --j) {
      int lo = last[j].first;
      if (lo < 0) break;
      cnt[k] += hi - lo;
      hi = lo;
      k++;
    }
    cnt[k] += hi + 1;
  }
  int total = 0;
  for (int i = 1; i <= 26; ++i)
    if (cnt[i] == 0)
      break;
    else
      total++;
  cout << total << endl;
  for (int i = 1; i <= 26; ++i) {
    if (cnt[i] == 0)
      break;
    else
      cout << cnt[i] << endl;
  }
  return EXIT_SUCCESS;
}