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 = 1000005;
char str[N];
long long ans[30];
long long cnt[26];
long long gao(int l) {
  long long ans = 0;
  memset(cnt, 0, sizeof(cnt));
  int n = strlen(str), cur = 0;
  for (int i = 0, j = 0; i < n; i++) {
    while (cur < l && j < n) {
      if (cnt[str[j] - 'a'] == 0) cur++;
      cnt[str[j++] - 'a']++;
    }
    if (cur == l) {
      ans += n - j + 1;
    }
    if ((--cnt[str[i] - 'a']) == 0) cur--;
  }
  return ans;
}
int main() {
  int ret = 26;
  cin >> str;
  for (int i = 1; i <= 26; i++) {
    ans[i] = gao(i);
    if (ans[i] == 0) {
      ret = i - 1;
      break;
    }
  }
  cout << ret << endl;
  for (int i = 1; i <= ret; i++) cout << ans[i] - ans[i + 1] << endl;
  return 0;
}