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;
struct rec {
  int idx;
  char ch;
} recent[100];
bool comp(rec a, rec b) { return a.idx < b.idx; }
int main() {
  ios_base::sync_with_stdio(0);
  int i, j;
  string s;
  cin >> s;
  int n = s.length();
  long long ans[30];
  for (i = 0; i < 26; i++) {
    recent[i].idx = -1;
    recent[i].ch = 'a' + i;
    ans[i] = 0;
  }
  ans[26] = 0;
  for (i = n - 1; i >= 0; i--) {
    int pos = i;
    int k = 1;
    for (j = 0; j < 26; j++) {
      if (recent[j].ch == s[i]) {
        recent[j].idx = i;
        continue;
      }
      if (recent[j].idx == -1) continue;
      ans[k++] += (recent[j].idx - pos);
      pos = recent[j].idx;
    }
    ans[k] += n - pos;
    sort(recent, recent + 26, comp);
  }
  set<char> st;
  for (i = 0; i < n && st.size() < 26; i++) st.insert(s[i]);
  printf("%d\n", st.size());
  for (i = 1; i <= st.size(); i++) printf("%I64d\n", ans[i]);
  return 0;
}