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;
struct _ {
  ios_base::Init i;
  _() {
    cin.sync_with_stdio(0);
    cin.tie(0);
  }
} _;
const int N = 3e5 + 7, M = 26;
long long pref[N][M + 1];
int cu[N];
long long ans[M + 1];
int main() {
  string s;
  cin >> s;
  int n = s.size();
  int prev[M];
  for (int i = 0; i < M; ++i) prev[i] = -1;
  for (int i = 0; i < N; ++i) cu[i] = 0;
  for (int i = 0; i < n; ++i) {
    int last = prev[s[i] - 'a'];
    prev[s[i] - 'a'] = i;
    if (last >= 0)
      for (int j = 1; j <= M; ++j) ans[j] += pref[last][j];
    long long ct1[M + 1], ct2[M + 1];
    for (int j = 1; j <= M; ++j) ct1[j] = ct2[j] = 0;
    for (int j = last + 1; j < i; ++j) {
      ++ct1[cu[j]];
      ++cu[j];
      ++ct2[cu[j]];
      ++ans[cu[j]];
      for (int k = 1; k <= M; ++k) pref[j][k] += ct2[k] - ct1[k];
    }
    if (i > 0)
      for (int k = 1; k <= M; ++k) pref[i][k] = pref[i - 1][k];
    ++pref[i][1];
    ++ans[1];
    cu[i] = 1;
  }
  set<char> ss;
  for (int i = 0; i < n; ++i) ss.insert(s[i]);
  int sz = ss.size();
  cout << sz << '\n';
  for (int i = 1; i <= sz; ++i) cout << ans[i] << '\n';
  return 0;
}