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;
const int MAX = 3e5 + 9;
string s;
int a[255], d, n, b[MAX];
int main() {
  cin >> s, n = s.size();
  for (auto i : s) a[i] = 1;
  for (int i = 0; i < 255; i++) d += a[i];
  cout << d << '\n';
  for (int i = 1; i <= d; i++) {
    memset(a, 0, sizeof a);
    int p1 = 0, p2 = 0, dd = 0;
    long long ans = 0;
    while (p1 < n) {
      while (p2 < n)
        if (a[s[p2]] || (a[s[p2]] == 0 && dd < i))
          a[s[p2]]++, dd += (a[s[p2]] == 1), p2++;
        else
          break;
      if (dd == i) ans += (p2 - p1) - b[p1], b[p1] = p2 - p1;
      a[s[p1]]--, dd -= (!a[s[p1]]), p1++;
    }
    cout << ans << '\n';
  }
}