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;
int main() {
  string s;
  while (cin >> s) {
    int len = s.length(), maxans = 0;
    long long ans[27];
    memset(ans, 0, sizeof ans);
    pair<int, int> mp[27];
    const int maxn = 999999999;
    for (int i = 0; i < 26; i++) mp[i].first = maxn, mp[i].second = i;
    mp[26].first = -1, mp[26].second = 26;
    for (int i = 0; i < len; i++) {
      for (int j = 0; j < 27; j++)
        if (mp[j].second == s[i] - 'a')
          maxans += (mp[j].first == maxn ? 1 : 0), mp[j].first = i;
      sort(mp, mp + 27);
      for (int j = maxans, k = 0; j > 0; j--, k++)
        ans[j] += mp[k + 1].first - mp[k].first;
    }
    cout << maxans << endl;
    for (int i = 1; i <= maxans; i++) cout << ans[i] << endl;
  }
  return 0;
}