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;
deque<int> que[30];
string str;
int main() {
  cin >> str;
  int n = str.length();
  vector<long long> counter;
  counter.push_back(0);
  for (int i = 0; i < n; i++) {
    que[str[i] - 'a'].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    vector<int> pos;
    for (int j = 0; j < 26; j++) {
      while (!que[j].empty() && que[j].front() < i) que[j].pop_front();
      if (!que[j].empty()) pos.push_back(que[j].front());
    }
    sort(pos.begin(), pos.end());
    pos.push_back(n);
    for (int j = 1; j < pos.size(); j++) {
      int t = pos[j] - pos[j - 1];
      if (counter.size() <= j)
        counter.push_back(t);
      else
        counter[j] += t;
    }
  }
  cout << counter.size() - 1 << endl;
  for (int i = 1; i < counter.size(); i++) {
    cout << counter[i] << endl;
  }
  return 0;
}