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 N = 300010;
int n;
int num[N];
int h1[27], h2[27];
long long ans[N];
char s[N];
int main() {
  int t, i, lim, num1, num2, l, r;
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (lim = 1; lim <= 26; ++lim) {
    memset(h1, 0, sizeof h1);
    memset(h2, 0, sizeof h2);
    num1 = num2 = 0;
    l = 1;
    r = 1;
    for (i = 1; i <= n; ++i) {
      if (!h1[s[i] - 'a']++) ++num1;
      if (!h2[s[i] - 'a']++) ++num2;
      while (num1 > lim)
        if (!--h1[s[l++] - 'a']) --num1;
      while (num2 >= lim)
        if (!--h2[s[r++] - 'a']) --num2;
      if (num1 >= lim) ans[lim] = ans[lim] + r - l;
    }
    if (!ans[lim]) break;
  }
  --lim;
  printf("%d\n", lim);
  for (i = 1; i <= lim; ++i) printf("%I64d\n", ans[i]);
  return 0;
}