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 ch[300100];
int last[33];
long long num_sub[33], ans[33];
int main() {
  scanf("%s", ch + 1);
  int len = strlen(ch + 1);
  int cnt_ch, seg_len, i, j, k;
  for (i = 1; i <= len; i++) {
    j = last[ch[i] - 'a'];
    seg_len = i - j - 1;
    for (cnt_ch = 1; seg_len; cnt_ch++) {
      seg_len -= num_sub[cnt_ch];
    }
    num_sub[cnt_ch] += num_sub[cnt_ch - 1];
    for (k = cnt_ch - 1; k >= 1; k--) {
      num_sub[k] = num_sub[k - 1];
    }
    num_sub[1]++;
    last[ch[i] - 'a'] = i;
    for (k = 1; k <= 26; k++) {
      ans[k] += num_sub[k];
    }
  }
  int cnt = 26;
  while (ans[cnt] == 0) cnt--;
  printf("%d\n", cnt);
  for (i = 1; i <= cnt; i++) printf("%lld\n", ans[i]);
  return 0;
}