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 = 300005;
string s;
long long ans[26];
int n, dp[26][maxn], cnt[26];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s;
  n = s.length();
  for (int d = 1; d < 27; ++d) {
    fill(cnt, cnt + 26, 0);
    int cur = 0;
    for (int i = 0, j = 0; i < n; ++i) {
      if (i > 0 && cnt[s[i - 1] - 'a']-- == 1) --cur;
      for (; j < n && cur < d; ++j)
        if (cnt[s[j] - 'a']++ == 0) ++cur;
      dp[d - 1][i] = cur == d ? j : (n + 1);
    }
  }
  int res = 0;
  while (res < 26 && dp[res][0] != n + 1) ++res;
  for (int d = 0; d < res; ++d)
    for (int i = 0; i < n; ++i)
      ans[d] += (d < res - 1 ? dp[d + 1][i] : (n + 1)) - dp[d][i];
  cout << res << '\n';
  for (int i = 0; i < res; ++i) cout << ans[i] << '\n';
  return 0;
}