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 ptr[27];
long long int freq[27][27];
long long int cnt[27];
long long int ans[27];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  for (int i = 0; i < s.length(); i++) {
    for (int j = 1; j <= 26; j++) {
      freq[j][s[i] - 'a']++;
      if (freq[j][s[i] - 'a'] == 1) cnt[j]++;
    }
    for (int j = 1; j <= 26; j++) {
      while (cnt[j] > j) {
        freq[j][s[ptr[j]] - 'a']--;
        if (freq[j][s[ptr[j]] - 'a'] == 0) cnt[j]--;
        ptr[j]++;
      }
    }
    ans[1] += i - ptr[1] + 1;
    for (int j = 2; j <= 26; j++) {
      if (cnt[j]) {
        ans[j] += ptr[j - 1] - ptr[j];
      }
    }
  }
  for (int i = 26; i >= 1; i--)
    if (ans[i]) {
      cout << i << '\n';
      break;
    }
  for (int i = 1; i <= 26 && ans[i]; i++) cout << ans[i] << '\n';
  return 0;
}