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 long long N = 3e5 + 5;
const long long mod = 1e9 + 7;
const long long inf = 1e18;
const long long X = 1e7;
int dx[] = {0, 0, -1, 1};
int dy[] = {1, -1, 0, 0};
long long c[N], f[N][26], d[30];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  string s;
  cin >> s;
  int n = s.size();
  s = '*' + s + '*';
  for (long long i = n; i >= 1; i--) {
    for (long long j = 0; j <= 25; j++) f[i][j] = f[i + 1][j];
    f[i][s[i] - 'a'] = i;
  }
  int mx = 0;
  for (long long i = 1; i <= n; i++) {
    int ind[27], cnt = 0;
    for (long long x = 0; x <= 25; x++) ind[x] = f[i][x];
    sort(ind + 0, ind + 25 + 1);
    ind[26] = n + 1;
    for (long long x = 0; x <= 25; x++)
      if (ind[x] != 0) {
        cnt += 1;
        d[cnt] += ind[x + 1] - ind[x];
      }
    mx = max(mx, cnt);
  }
  cout << mx << "\n";
  for (long long i = 1; i <= mx; i++) cout << d[i] << "\n";
}