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;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
int n;
int m;
long long cnt[27];
int last[26];
int g[300000][26];
string s;
int main() {
  cin >> s;
  n = ((int)(s).size());
  for (int i = 0; i < 26; i++) last[i] = n;
  for (int i = n - 1; i >= 0; i--) {
    last[s[i] - 'a'] = i;
    for (int j = 0; j < 26; j++) g[i][j] = last[j];
  }
  for (int i = 0; i < n; i++) {
    int mask = 1 << (s[i] - 'a'), cur = 1;
    int j = i;
    while (j < n) {
      int k = n, ch = 0;
      for (int t = 0; t < 26; t++)
        if (((mask >> t) & 1) == 0 && g[j][t] < k) {
          k = g[j][t];
          ch = t;
        }
      cnt[cur] += k - j;
      cur++;
      mask |= 1 << ch;
      j = k;
    }
  }
  int ans = 0;
  while (ans < 26 && cnt[ans + 1]) ans++;
  printf("%d\n", ans);
  for (int i = 1; i <= 26; i++)
    if (cnt[i] != 0) printf("%I64d\n", cnt[i]);
  return 0;
}