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;
const int MAXN = 300100;
int N;
string str;
int s[MAXN];
long long res[MAXN];
int last[26];
vector<int> t;
int main() {
  cin >> str;
  N = str.length();
  for (int i = 0; i < N; i++) s[i] = str[i] - 'a';
  for (int i = 0; i < N; i++) res[i] = 0;
  for (int i = 0; i < 26; i++) last[i] = 1e9;
  for (int i = N - 1; i >= 0; i--) {
    last[s[i]] = i;
    t.clear();
    for (int j = 0; j < 26; j++)
      if (last[j] < N) t.push_back(last[j]);
    sort(t.begin(), t.end());
    t.push_back(N);
    for (int j = 1; j < t.size(); j++) res[j] += (t[j] - t[j - 1]);
  }
  int ans = 1;
  while (res[ans]) ans++;
  cout << ans - 1 << "\n";
  for (int i = 1; i < ans; i++) cout << res[i] << "\n";
  return 0;
}