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>
const int maxn = 310000;
long long f[maxn + 1];
int K, n;
char s[maxn + 1];
int ge[26];
void init() {
  int i;
  bool you[26];
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (i = 0; i < 26; ++i) you[i] = false;
  for (i = 1; i <= n; ++i) you[s[i] - 'a'] = true;
  K = 0;
  for (i = 0; i < 26; ++i)
    if (you[i]) ++K;
}
void work() {
  int i, j, yao, diff;
  int ge[26];
  for (yao = 1; yao <= K; ++yao) {
    for (j = 0; j < 26; ++j) ge[j] = 0;
    j = 1;
    diff = 0;
    for (i = 1; i <= n; ++i) {
      if (i > 1) {
        --ge[s[i - 1] - 'a'];
        if (ge[s[i - 1] - 'a'] == 0) --diff;
      }
      while (j <= n) {
        ++ge[s[j] - 'a'];
        if (ge[s[j] - 'a'] == 1) ++diff;
        if (diff > yao) {
          --ge[s[j] - 'a'];
          --diff;
          break;
        }
        ++j;
      }
      f[yao] += j - i;
    }
  }
  for (i = K; i >= 2; --i) f[i] -= f[i - 1];
}
void output() {
  int i;
  printf("%d\n", K);
  for (i = 1; i <= K; ++i) printf("%I64d\n", f[i]);
}
int main() {
  init();
  work();
  output();
  return 0;
}