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() {
  string s;
  cin >> s;
  int n = s.length();
  int lst[26];
  bool hs[26] = {false};
  int a[n];
  for (int i = 0; i < n; i++) {
    if (hs[s[i] - 'a'])
      a[i] = lst[s[i] - 'a'];
    else {
      a[i] = -1;
      hs[s[i] - 'a'] = true;
    }
    lst[s[i] - 'a'] = i;
  }
  int sum = 0;
  for (int j = 0; j < 26; j++) sum += hs[j];
  cout << sum << endl;
  vector<int> nd[n];
  nd[0].push_back(0);
  for (int i = 1; i < n; i++) {
    nd[i] = nd[i - 1];
    nd[i].push_back(i);
    if (a[i] != -1) nd[i].erase(upper_bound(nd[i].begin(), nd[i].end(), a[i]));
  }
  unsigned long long dv[26] = {0ULL};
  for (int i = 0; i < n; i++) dv[0] += (i + 1 - nd[i].back());
  for (int i = 0; i < n; i++)
    for (auto it = nd[i].rbegin(); next(it) != nd[i].rend(); it++)
      dv[1 + (it - nd[i].rbegin())] += (*it - *next(it));
  for (int i = 0; i < sum; i++) cout << dv[i] << endl;
  return 0;
}