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;
long long const MAXN = 3e5 + 10;
long long n, m, T;
string s;
long long pos[30], p[30], ans[30];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> s;
  memset(pos, -1, sizeof pos);
  n = s.size();
  for (long long i = 0; i < n; i++) {
    pos[s[i] - 'a'] = i;
    long long cnt = 0;
    for (long long c = 0; c < 26; c++) {
      if (pos[c] != -1) p[cnt++] = pos[c];
    }
    sort(p, p + cnt);
    for (long long j = 1; j < cnt; j++) {
      ans[cnt - j] += p[j] - p[j - 1];
    }
    ans[cnt] += p[0] + 1;
  }
  for (long long i = 26; i >= 1; i--) {
    if (ans[i]) {
      cout << i << endl;
      for (long long j = 1; j <= i; j++) cout << ans[j] << endl;
      break;
    }
  }
  return 0;
}