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 int M = 998244353;
string s;
long long int l;
set<long long int, greater<long long int>> sett;
long long int ans[27] = {0};
void update(long long int i) {
  auto it = sett.begin();
  long long int cc = 1;
  while (it != sett.end()) {
    ans[cc] += i - *it;
    i = *it;
    cc++;
    it++;
  }
}
int main() {
  cin >> s;
  l = s.length();
  s = '`' + s;
  long long int index[27];
  memset(index, 0, sizeof(index));
  for (long long int i = 1; i <= l; i++) {
    long long int c = s[i] - 'a' + 1;
    sett.insert(index[s[i - 1] - 'a' + 1]);
    if (index[c]) sett.erase(index[c]);
    index[c] = i;
    update(i);
  }
  long long int ds = 26;
  for (long long int i = 1; i < 27; i++)
    if (ans[i] == 0) {
      ds = i - 1;
      break;
    }
  cout << ds << endl;
  for (long long int i = 1; i <= ds; i++) cout << ans[i] << endl;
  return 0;
}