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 diversity(string s) {
  set<char> letters;
  for (char l : s) {
    letters.insert(l);
  }
  return letters.size();
}
long long count(string s, int d) {
  int n = s.length();
  int occ[26];
  for (int i = 0; i < 26; i++) {
    occ[i] = 0;
  }
  long long result = 0;
  int e1 = -1, e2 = -1, curr_d = 0;
  for (int p = 0; p < n; p++) {
    while ((e1 < p) || (e1 < n - 1 && curr_d < d)) {
      e1++;
      int chr = s[e1] - 'a';
      occ[chr]++;
      if (occ[chr] == 1) {
        curr_d++;
      }
    }
    while ((e2 < e1) || (e2 < n - 1 && occ[s[e2 + 1] - 'a'] > 0)) {
      e2++;
    }
    if (curr_d == d) {
      result += e2 - e1 + 1;
    }
    occ[s[p] - 'a']--;
    if (occ[s[p] - 'a'] == 0) {
      curr_d--;
    }
  }
  return result;
}
int main() {
  string s;
  cin >> s;
  int d = diversity(s);
  cout << d << endl;
  for (int i = 1; i <= d; i++) {
    cout << count(s, i) << endl;
  }
  return 0;
}