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 L;
char s[300005];
int nxt[300005][26];
long long ans[300005];
int main() {
  scanf("%s", s);
  L = strlen(s);
  for (int i = 0; i < 26; ++i) nxt[L][i] = L;
  for (int i = L - 1; i >= 0; --i) {
    for (int j = 0; j < 26; ++j) nxt[i][j] = nxt[i + 1][j];
    nxt[i][s[i] - 'a'] = i;
  }
  memset(ans, 0, sizeof ans);
  int D = 0;
  for (int i = 0; i < L; ++i) {
    int mask = (1 << (s[i] - 'a')), d = 0, pos = i;
    while (pos < L) {
      int e = L;
      for (int j = 0; j < 26; ++j)
        if ((mask & (1 << j)) == 0) e = min(e, nxt[i + 1][j]);
      ans[++d] += e - pos;
      pos = e;
      for (int j = 0; j < 26; ++j)
        if (nxt[i + 1][j] == e) mask |= (1 << j);
    }
    D = max(D, d);
  }
  printf("%d\n", D);
  for (int i = 1; i <= D; ++i) printf("%I64d\n", ans[i]);
  return 0;
}