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;
static char s[500000];
static long long m1[26], m2[26];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s;
  long long n = strlen(s), m = 0, a, b, r, c;
  for (long long i = 0; i < n; ++i) {
    m1[s[i] - 'a']++;
  }
  for (long long i = 0; i < 26; ++i)
    if (m1[i]) ++m;
  memset(m1, 0, sizeof(m1));
  cout << m << endl;
  for (long long i = 1; i <= m; ++i) {
    memset(m1, 0, sizeof(m1));
    a = -1;
    b = 0;
    while (a < n && b < i) {
      ++a;
      m1[s[a] - 'a']++;
      if (m1[s[a] - 'a'] == 1) ++b;
    }
    m2[i] = n - a;
    for (long long j = 0; a < n; ++j) {
      if (m1[s[j] - 'a'] == 1) --b;
      m1[s[j] - 'a']--;
      while (a < n && b < i) {
        ++a;
        m1[s[a] - 'a']++;
        if (m1[s[a] - 'a'] == 1) ++b;
      }
      m2[i] += n - a;
    }
  }
  for (long long i = 1; i < m; ++i) cout << m2[i] - m2[i + 1] << endl;
  cout << m2[m] << endl;
  return 0;
}