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 N = 300001;
long long dp[N][27], cnt[27];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string s;
  cin >> s;
  int ds = 0, n = s.size();
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j < 26; ++j) {
      if (s[i - 1] == j + 'a')
        dp[i][j] = i;
      else
        dp[i][j] = dp[i - 1][j];
    }
  for (int i = 1; i <= n; ++i) {
    vector<int> z;
    for (int j = 0; j < 26; ++j)
      if (dp[i][j]) z.push_back(dp[i][j]);
    int d = z.size();
    if (i == n) ds = d;
    z.push_back(0);
    sort(z.begin(), z.end());
    for (int j = 1; j < z.size(); ++j, --d) cnt[d] += z[j] - z[j - 1];
  }
  cout << ds << "\n";
  for (int i = 1; i <= ds; ++i) cout << cnt[i] << "\n";
}