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 K = 26;
int first[K];
long long ans[K + 1];
int main() {
  for (int i = 0; i < K; i++) first[i] = -1;
  string s;
  cin >> s;
  vector<int> now;
  for (int i = 0; i < s.size(); i++) {
    int c = s[i] - 'a';
    vector<int> tmp(1, i);
    for (int j = 0; j < now.size(); j++)
      if (first[c] != now[j]) tmp.push_back(now[j]);
    sort(tmp.begin(), tmp.end());
    now = tmp;
    first[c] = i;
    for (int j = tmp.size() - 1; j >= 0; j--)
      ans[tmp.size() - j] += tmp[j] - (j ? tmp[j - 1] : -1);
  }
  int sz = K;
  for (int i = 1; i <= K; i++)
    if (!ans[i]) {
      sz = i - 1;
      break;
    }
  cout << sz << endl;
  for (int i = 1; i <= sz; i++) cout << ans[i] << endl;
  return 0;
}