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;
int main() {
  int timeOfUse[26] = {0};
  long long numOfSubstringWithDiversity[27] = {0};
  char a[300010];
  cin.getline(a, 300010, '\n');
  int countDiversities = 0;
  for (int i = 0; i < strlen(a); i++) {
    if (timeOfUse[a[i] - 'a'] == 0) {
      timeOfUse[a[i] - 'a']++;
      countDiversities++;
    }
  }
  int m = countDiversities;
  cout << countDiversities << endl;
  for (int diversity = 1; diversity <= 26; diversity++) {
    memset(timeOfUse, 0, sizeof timeOfUse);
    countDiversities = 0;
    for (int i = 0, j = 0; i < strlen(a); i++) {
      if (timeOfUse[a[i] - 'a'] == 0) countDiversities++;
      timeOfUse[a[i] - 'a']++;
      for (; countDiversities > diversity; j++) {
        timeOfUse[a[j] - 'a']--;
        if (timeOfUse[a[j] - 'a'] == 0) countDiversities--;
      }
      numOfSubstringWithDiversity[diversity] += i - j + 1;
    }
  }
  for (int i = 1; i <= m; i++) {
    cout << numOfSubstringWithDiversity[i] - numOfSubstringWithDiversity[i - 1]
         << endl;
  }
  return 0;
}