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 str[300001];
long long ans[28];
int cnt[28], flag[28], tt[28], c;
int main() {
  int i, j, t;
  scanf("%s", str);
  t = c = 0;
  for (i = 0; i < 28; i++) tt[i] = 0;
  for (i = 0; i < strlen(str); i++) {
    if (tt[str[i] - 'a'] == 0) c++, tt[str[i] - 'a'] = 1;
  }
  for (i = 0; i < strlen(str); i++) {
    if (flag[str[i] - 'a']) {
      cnt[flag[str[i] - 'a']] += cnt[flag[str[i] - 'a'] - 1];
    }
    int tmp = flag[str[i] - 'a'];
    for (j = c + 1; j >= 1; j--) {
      if (tmp) {
        if (j < tmp) cnt[j] = cnt[j - 1];
      } else
        cnt[j] = cnt[j - 1];
      if (j != 1) ans[j] += cnt[j];
      if ((!tmp && flag[j - 1]) || (flag[j - 1] && flag[j - 1] < tmp))
        flag[j - 1]++;
    }
    if (!tmp) t++;
    if (tmp == 1)
      cnt[1]++;
    else
      cnt[1] = 1;
    flag[str[i] - 'a'] = 1;
    ans[1] += cnt[1];
  }
  printf("%d\n", t);
  for (i = 1; i <= t; i++) printf("%I64d\n", ans[i]);
  return 0;
}