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 = 3 * 100 * 1000 + 100, P = 27;
long long n, a[P], ds, b[P], ans[P];
string s;
bool mark[P];
vector<long long> v[P];
int main() {
  cin >> s;
  n = s.size();
  memset(b, -1, sizeof b);
  for (int i = 0; i < n; i++) {
    mark[s[i] - 'a'] = true;
    v[s[i] - 'a'].push_back(i);
    b[s[i] - 'a'] = 0;
  }
  for (int i = 0; i < P; i++)
    if (mark[i]) ds++;
  for (int i = 0; i < n; i++) {
    long long cnt = 0;
    for (int j = 0; j < P; j++)
      if (b[j] != -1) {
        a[cnt++] = v[j][b[j]];
      }
    sort(a, a + cnt);
    a[cnt] = n;
    for (int j = 1; j <= cnt; j++) {
      ans[j] += a[j] - a[j - 1];
    }
    for (int j = 0; j < P; j++)
      if (b[j] != -1 && v[j][b[j]] == i) {
        if (b[j] == v[j].size() - 1)
          b[j] = -1;
        else
          b[j]++;
      }
  }
  cout << ds << endl;
  for (int i = 1; i <= ds; i++) cout << ans[i] << endl;
  return 0;
}