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 int64_t N = 3 * 1e5 + 2;
int main() {
  char sc[N];
  scanf("%s", sc);
  string s(sc);
  int64_t n = s.size();
  vector<vector<int64_t>> index(27);
  vector<vector<int64_t>> dp2(n + 1);
  for (int64_t i = 0; i < s.size(); i++) {
    index[s[i] - 'a'].push_back(i);
    for (char c = 'a'; c <= 'z'; c++)
      if (!index[c - 'a'].empty()) dp2[i].push_back({*index[c - 'a'].rbegin()});
  }
  vector<int64_t> dp(27, 0);
  dp[1] = 1;
  int64_t maxd = dp2[n - 1].size();
  for (int64_t i = 1; i < s.size(); i++) {
    sort(dp2[i].begin(), dp2[i].end(), std::greater<int64_t>{});
    auto& ss = dp2[i];
    ss.push_back(-1);
    for (int64_t j = 0; j < ss.size() - 1; j++)
      dp[j + 1] += (ss[j] - ss[j + 1]);
  }
  cout << maxd << endl;
  for (int64_t i = 1; i <= maxd; i++) cout << dp[i] << endl;
}