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;
string s;
long long int ans[100001], cnt[500];
long long int solve(long long int x) {
  long long int r = 0;
  long long int came = 0;
  long long int res = 0;
  memset(cnt, 0, sizeof cnt);
  for (long long int l = 0; l < s.length(); l++) {
    while (r < s.length() && came <= x) {
      cnt[s[r] - 'a']++;
      if (cnt[s[r] - 'a'] == 1) came++;
      r++;
    }
    if (came <= x)
      res += r - l;
    else
      res += r - l - 1;
    cnt[s[l] - 'a']--;
    if (cnt[s[l] - 'a'] == 0) came--;
  }
  return res;
}
int main() {
  long long int i, j, maxi;
  cin >> s;
  maxi = -1;
  for (i = 26; i >= 0; i--) {
    ans[i] = solve(i);
    ans[i + 1] -= ans[i];
    if (ans[i + 1] > 0) maxi = max(maxi, i + 1);
  }
  cout << maxi << endl;
  for (i = 1; i <= maxi; i++) cout << ans[i] << endl;
  return 0;
}