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;
char str[300005];
vector<int> idx[27];
int main() {
  int x;
  scanf("%s", str);
  int n = strlen(str);
  for (int i = 0; i < (int)(n); i++) {
    idx[str[i] - 'a'].push_back(i);
  }
  long long int ans[27] = {0};
  int ansmax = 0;
  for (int i = 0; i < (int)(n); i++) {
    set<int> fir;
    for (int j = 0; j < (int)(26); j++) {
      vector<int>::iterator it = lower_bound(idx[j].begin(), idx[j].end(), i);
      if (it != idx[j].end()) fir.insert(*it);
    }
    int p = i;
    int c = 0;
    for (__typeof((fir).begin()) it = (fir).begin(); it != (fir).end(); ++it) {
      ans[c++] += *it - p;
      p = *it;
    }
    ans[c] += n - p;
    ansmax = max(ansmax, c);
  }
  cout << ansmax << endl;
  for (int i = 0; i < (int)(ansmax); i++) cout << ans[i + 1] << endl;
  return 0;
}