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;
void dout() { cerr << endl; }
template <typename Head, typename... Tail>
void dout(Head H, Tail... T) {
  cerr << H << ' ';
  dout(T...);
}
char s[500000];
pair<int, int> y[500000];
vector<long long> arr;
int main() {
  scanf("%s", s);
  int len = strlen(s);
  for (int i = 0; i < len; ++i) y[i].first = i, y[i].second = 0;
  for (int it = 1; it <= 26; ++it) {
    long long ans = 0;
    for (int i = 0, pos = 0, lastadd = 0, l = 0; i < len; ++i) {
      if (i > 0 && y[i].first == l) {
        ans += lastadd;
        y[i].first = y[i - 1].first, y[i].second = y[i - 1].second;
        continue;
      }
      if (y[i].first == len) break;
      l = y[i].first;
      y[i].second |= (1 << (s[y[i].first] - 'a'));
      while (pos < len && ((y[i].second & (1 << (s[pos] - 'a'))) != 0)) ++pos;
      ans += pos - l, lastadd = pos - l;
      y[i].first = pos;
    }
    if (ans == 0) break;
    arr.push_back(ans);
  }
  printf("%d\n", arr.size());
  for (int i = 0; i < arr.size(); ++i) printf("%I64d\n", arr[i]);
  return 0;
}