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;
char s[300005], v[300005];
int ap[300005][26], pos[27], len[26], n;
long long sol[27];
void solve(int idx) {
  int nr = 0;
  for (int i = 0; i < 26; ++i) {
    int st = 1, dr = len[i], ans = n + 1;
    while (st <= dr) {
      int mij = (st + dr) / 2;
      if (ap[mij][i] >= idx)
        ans = ap[mij][i], dr = mij - 1;
      else
        st = mij + 1;
    }
    if (ans <= n) ++nr, pos[nr] = ans;
  }
  sort(pos + 1, pos + nr + 1);
  for (int i = 1; i <= nr; ++i)
    if (i < nr)
      sol[i] += pos[i + 1] - pos[i];
    else
      sol[i] += n - pos[i] + 1;
}
int main() {
  int i, nr = 0;
  cin.get(s, 300005);
  cin.get();
  n = strlen(s);
  for (i = 1; i <= n; ++i)
    v[i] = s[i - 1], ++len[v[i] - 'a'], ap[len[v[i] - 'a']][v[i] - 'a'] = i;
  for (i = 0; i < 26; ++i)
    if (len[i] > 0) ++nr;
  for (i = 1; i <= n; ++i) solve(i);
  cout << nr << '\n';
  for (i = 1; i <= nr; ++i) cout << sol[i] << '\n';
  return 0;
}