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;
int fs[26], now[26];
char str[300005];
long long int cnt[300005];
int main() {
  scanf("%s", &str);
  int n = strlen(str);
  for (int i = 0; i < 26; i++) fs[i] = n;
  for (int i = n - 1; i >= 0; i--) {
    fs[str[i] - 'a'] = i;
    for (int j = 0; j < 26; j++) now[j] = fs[j];
    sort(now, now + 26);
    for (int j = 0; j < 26; j++) {
      if (j == 26 - 1)
        cnt[j] += (long long int)n - now[j];
      else
        cnt[j] += (long long int)now[j + 1] - now[j];
    }
  }
  int lim = 0;
  for (int i = 0; i < 26; i++) {
    if (cnt[i] == 0) break;
    lim = i;
  }
  printf("%d\n", lim + 1);
  for (int j = 0; j <= lim; j++) printf("%I64d\n", cnt[j]);
  return 0;
}