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;
const int MAXN = 300100;
string s;
int n, freq[200];
long long ans[30];
set<char> st;
long long go(int sze) {
  int lt = 0, rt = 0, cnt = 0;
  long long ret = 0;
  memset(freq, 0, sizeof(freq));
  while (lt < n) {
    if (rt == n || cnt == sze + 1) {
      if (cnt == sze + 1)
        ret += rt - lt - 1;
      else
        ret += rt - lt;
      freq[s[lt]]--;
      if (freq[s[lt]] == 0) cnt--;
      lt++;
    } else {
      freq[s[rt]]++;
      if (freq[s[rt]] == 1) cnt++;
      rt++;
    }
  }
  return ret;
}
int main() {
  ios ::sync_with_stdio(false);
  cin >> s;
  n = s.length();
  for (int i = 0; i < s.length(); i++) st.insert(s[i]);
  cout << st.size() << '\n';
  for (int i = 1; i <= st.size(); i++) ans[i] = go(i);
  for (int i = 1; i <= st.size(); i++) cout << ans[i] - ans[i - 1] << '\n';
  return 0;
}