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;
class node {
 public:
  node(char v, int p) { val = v, pos = p; }
  char val;
  int pos;
};
int main(void) {
  char c;
  unsigned long long cnt[26] = {0};
  int i = 0, j = 0;
  bool found;
  list<node> h;
  int dis = 0;
  while ((c = getchar()) != '\n') {
    cnt[0]++;
    found = false;
    j = 0;
    for (list<node>::iterator it = h.begin(); it != h.end(); j++) {
      list<node>::iterator next = it;
      next++;
      if (next == h.end())
        dis = it->pos + 1;
      else
        dis = it->pos - next->pos;
      if (found)
        cnt[j] += dis;
      else if (it->val != c)
        cnt[j + 1] += dis;
      else {
        found = true;
        cnt[j] += dis;
        it = h.erase(it);
        h.push_front(node(c, i));
        continue;
      }
      it++;
    }
    if (!found) {
      h.push_front(node(c, i));
    }
    i++;
  }
  int hl = h.size();
  printf("%d\n", hl);
  for (int i = 0; i < hl; i++) cout << cnt[i] << endl;
  ;
  return 0;
}