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 ts = "abcdefghijklmnopqrstuvwxyz";
int main() {
  string s;
  cin >> s;
  int n = (int)s.size();
  int p[26];
  for (int i = 0; i < 26; ++i) {
    p[i] = (int)s.find(ts.substr(i, 1));
  }
  long long ans[27] = {};
  for (int i = 0; i < n; ++i) {
    int tmp[27];
    int k = 0;
    for (int c = 0; c < 26; ++c) {
      if (p[c] >= i) {
        tmp[k++] = p[c];
      }
    }
    tmp[k++] = n;
    sort(tmp, tmp + k);
    for (int c = 1; c < k; ++c) {
      ans[c] += tmp[c] - tmp[c - 1];
    }
    p[s[i] - 'a'] = (int)s.find(s.substr(i, 1), i + 1);
  }
  int cnt = 1;
  while (cnt < 26 && ans[cnt + 1] > 0) {
    ++cnt;
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; ++i) {
    cout << ans[i] << endl;
  }
}