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;
char s[300001];
long long ts[27] = {};
pair<char, int> lasts[27];
int main() {
  gets(s);
  for (int i = 0; i < 27; i++) {
    lasts[i] = make_pair('a' + i, -1);
  }
  class _Pred {
   public:
    bool operator()(const pair<char, int>& a, const pair<char, int>& b) {
      return a.second > b.second;
    }
  } pred;
  int p = 0;
  while (s[p] != '\0') {
    sort(lasts, lasts + 26, pred);
    int last_p = p;
    int d = 0;
    for (int i = 0; i <= 26 && last_p != -1; i++) {
      if (lasts[i].first == s[p]) {
        continue;
      }
      ts[d] += last_p - lasts[i].second;
      last_p = lasts[i].second;
      d++;
    }
    for (int i = 0; i < 26; i++) {
      if (lasts[i].first == s[p]) {
        lasts[i].second = p;
      }
    }
    p++;
  }
  int d = 27;
  while (ts[--d] == 0)
    ;
  printf("%d\n", d + 1);
  for (int i = 0; i <= d; i++) {
    printf("%I64d\n", ts[i]);
  }
  return 0;
}