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 long long MOD = 1e9 + 7;
const long long INF = 1e18;
const string nl = "\n";
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  string second;
  cin >> second;
  set<char> se(second.begin(), second.end());
  cout << ((long long)(se).size()) << nl;
  long long n = ((long long)(second).size());
  auto get_ans = [&](long long x) -> long long {
    long long res = 0, r = 0;
    map<char, long long> cnt;
    for (long long i = 0; i < n; ++i) {
      while (r < n && ((long long)(cnt).size()) <= x) {
        ++cnt[second[r]];
        ++r;
      }
      res += (((long long)(cnt).size()) <= x ? r - i : r - i - 1);
      --cnt[second[i]];
      if (cnt[second[i]] == 0) {
        cnt.erase(second[i]);
      }
    }
    return res;
  };
  vector<long long> ans(((long long)(se).size()) + 5, 0LL);
  for (long long i = 1; i <= ((long long)(se).size()); ++i) {
    ans[i] = get_ans(i);
  }
  for (long long i = ((long long)(se).size()); i > 0; --i) {
    ans[i] -= ans[i - 1];
  }
  for (long long i = 1; i <= ((long long)(se).size()); ++i) {
    cout << ans[i] << nl;
  }
}