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 N = 3e5 + 10;
char second[N];
long long ans[30], res[30];
int main() {
  scanf("%s", second);
  int n = strlen(second);
  for (int t = 1; t <= 26; ++t) {
    int cnt[30] = {0}, diff = 0;
    for (int i = 0, j = -1; i < n; ++i) {
      while (j + 1 < n) {
        int c = second[j + 1] - 'a';
        cnt[c]++;
        if (cnt[c] == 1) diff++;
        if (diff > t) {
          cnt[c]--;
          diff--;
          break;
        }
        ++j;
      }
      ans[t] += j - i + 1;
      cnt[second[i] - 'a']--;
      if (cnt[second[i] - 'a'] == 0) --diff;
    }
  }
  for (int i = 1; i <= 26; ++i) res[i] = ans[i] - ans[i - 1];
  int e = 26;
  while (!res[e]) --e;
  cout << e << endl;
  for (int i = 1; i <= e; ++i) cout << res[i] << endl;
  return 0;
}