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;
long long int ans[27];
int ct[26];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  string s;
  cin >> s;
  memset(ans, 0, sizeof ans);
  int n = s.size();
  for (int sz = 1; sz <= 26; ++sz) {
    int p_c = 0;
    int j = 0;
    memset(ct, 0, sizeof ct);
    for (int i = 0; i < n; ++i) {
      if (ct[s[i] - 'a'] == 0) {
        ++p_c;
      }
      ++ct[s[i] - 'a'];
      while (p_c > sz) {
        --ct[s[j] - 'a'];
        if (ct[s[j] - 'a'] == 0) {
          --p_c;
        }
        ++j;
      }
      ans[sz] += i - j + 1;
    }
  }
  vector<long long int> f_v{};
  int mm = 26;
  for (int i = 1; i <= 26; ++i) {
    f_v.push_back(ans[i] - ans[i - 1]);
    if (f_v.back() == 0) {
      mm = i - 1;
      break;
    }
  }
  cout << mm << endl;
  for (int i = 1; i <= mm; ++i) {
    cout << f_v[i - 1] << endl;
  }
  return 0;
}