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 MAX = 400000 + 10;
int n;
char str[MAX];
int near[MAX];
long long ans[MAX];
int main() {
  scanf("%s", str + 1);
  n = strlen(str + 1);
  int i;
  set<int> num;
  for ((i) = (1); (i) <= (n); ++(i)) {
    if (near[str[i] - 'a']) num.erase(near[str[i] - 'a']);
    num.insert(near[str[i] - 'a'] = i);
    set<int>::iterator it;
    int last = 0, res = num.size();
    for ((it) = (num.begin()); (it) != (num.end()); ++(it)) {
      ans[res] += *it - last;
      last = *it;
      res--;
    }
  }
  for ((i) = (1); (i) <= (n); ++(i))
    if (!ans[i]) break;
  printf("%d\n", i - 1);
  for ((i) = (1); (i) <= (n); ++(i)) {
    if (!ans[i]) break;
    cout << ans[i] << endl;
  }
  return 0;
}